Calculus 2 Equations: Expert Solutions and More

I am trying to understand implicit differentiation; I understand what to do (that is no problem), but why I do it is another story. For example:
$3y^2=5x^3$
I understand that, if I take the derivative with respect to x of both sides of the equation, I'll get:
$\frac{d}{dx}(3y^2)=\frac{d}{dx}(5x^3)$
$6y\frac{d}{dx}(y)=15x^2\frac{d}{dx}(x)$
$6y\frac{dy}{dx}=15x^2\frac{dx}{dx}$
$6y\frac{dy}{dx}=15x^2$
$\frac{dy}{dx}=\frac{15x^2}{6y}$
Unless I made some sort of error, this is what I am suppose to do. But why? Specifically, on the second line, I utilize the chain rule for the "outer function" and get 6y, but I still need to utilize the chain rule for the "inner function" which is the y. So why don't I go ahead and take the derivative of y and get 1? I know that I am not suppose to, but I don't really "get it." It seems to me that I only use the chain rule "halfway". Why isn't it an all or nothing? If it's all done with respect to x, it would seem to me that the 3y^2 should remain unchanged entirely. This is my problem.

How do you show that ${\displaystyle f(x,y)=x^4+y^4}$ has a saddle point at (x,y) = (0,0)?

How do you differentiate ${\displaystyle y\sin x=y}$

Let $f:[0,1]\to <!--\; \to \; -->\mathbb{R}$ continuous and let $f([0,1])\subset <!--\; \subset \; -->[0,1]$. By using intermediate value theorem, show that $f$ has at least got one fixed point in $[0,1]$ (aka there is one solution $x\in <!--\; \in \; -->[0,1]$ of the equation $f(x)=x$).

I'm not sure at all if I did it right but I have taken the function:
$f(x)=x$
Then take the given intervals and insert for $x$ the beginning of interval:
$f(0)=0$
And the end of the interval:
$f(1)=1>0$
Because the function is continuous and because the equality sign changes to an inequality sign right after, the intermediate value theorem provides there must be at least one solution $x_1=[0,1]$.
Did I do it correctly? Did I explain correctly?

differential equations and physical intuition
Often when you study differential equations, you find phenomena in nature modeled by those equations. Sometimes an insight into a physical problem can help you to solve a differential equation. My question is: If you are a pure mathematician studying differential equations, do you have to be good at physics (biology, finance) too?

Use Implicit differentiation to prove
$y^{″}=\frac{-<!--\; -\; -->9}{y^2}$ if $4x^2-<!--\; -\; -->2y^2=9$

Say$f(x)=\ln <!--\; \; -->(1+2x+2x^2)$ or $g(x)=\tan <!--\; \; -->(2x^4-<!--\; -\; -->x)$.Using the definition leads to messy derivatives almost immediately.If it was some simple rational function,for example,i would try to use Maclaurin Series of $\frac{1}{1+x}$ or $\frac{1}{1-<!--\; -\; -->x}$ and then manipulate it to get result

Differential equations system
I have found the following example in one of my courses but I don't have any similar exercises resolved so I would like to know how to solve this:
The differential equations system is the following:
$$x_1^{\prime }=3x_1-<!--\; -\; -->2x_2+e^t$$
$$x_2^{\prime }=2x_1-<!--\; -\; -->x_2+2e^{2t}$$
a) Write the system in the matriceal form $x^{\prime }=Ax+b(t)$
b) Determinate the solutin of the system.

Division of differential equations
$$\frac{dx(t)}{dy(t)}=\frac{\mathrm{\alpha <!--\; \alpha \; -->}x(t)-<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}x(t)y(t)}{-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}y(t)+\mathrm{\delta <!--\; \delta \; -->}x(t)y(t)}$$
How would one simplify this fraction? Maybe the chain rule could be of any use, but I don't see how.

Use a table of integrals to evaluate the following integrals. $$\int <!--\; \int \; -->x^2{e}^{5x}dx$$

Differential equations notation
I've always wondered why does the differential equation notation for linear equations differ from the standard terminology of vector spaces.
We all know that the equation $y^{″}+p(x)y^{\prime }+q(x)y=g(x)$ for some function g is called linear and that the associated equation $y^{″}+p(x)y^{\prime }+q(x)y=0$ is called homogeneous. But why is that? WHY should mathematicians explicitly cause confusion with the rest of the theory of vector spaces?
What I mean by that is : Why not call the equation $y^{″}+p(x)y^{\prime }+q(x)y=g(x)$ an affine equation and call $y^{\prime }+p(x)y^{\prime }+q(x)y=0$ a linear equation? Because linear equations (in the sense of differential equations) are not linear in the sense of vector spaces unless they're homogeneous ; and linear equations (in the sense of differential equations) remind me more of a linear system of the form $Ax=b$ (which is called an affine equation in vector space theory) than of a linear equation at all.
Just so that I made myself clear ; I perfectly know the difference between linear equations in linear algebra and linear equations in differential equations theory ; I'm asking for some reason of "why the name".

using standard Maclaurin series (e.g. $e^x=1+x+\frac{x^2}{2!}+...$ , we can substitute $x$ with other functions such as $z=x^2$ to transform the standard Maclaurin series into $e^{x^2}=1+x^2+\frac{x^4}{2!}+...$ Why doesn't this require a $\frac{dz}{dx}=2x$ component in the expansion since the standard Maclaurin series are derived from differentiation of the original term.

If $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ is continuous over the entire domain Ms is not bounded above or below, show that $f(\mathbb{R})=\mathbb{R}.$

My approach would be to say that since $f$ is continuous on $\mathbb{R}$ it must be continuous on $[a,b],a,b\in <!--\; \in \; -->\mathbb{R}$ and using the intermediate value theorem this tells us that $f$ must take every value between $f(a)$ and $f(b)$. Now I want to argue that we can make this closed bounded interval as large as we want and the result still holds and I want to show that this implies that we can make the values of $f(a)$ as small as we like and $f(b)$ as large as we like (or the other way round) since $f$ is neither bounded below or above but I'm not sure if this argument can used to show that the image of $f$ is $\mathbb{R}$ since we switch from closed bounded intervals to an unbounded interval.

Does my argument hold to show the result?

I'm sure this question has been asked before and yet I can't find it.
Is there a proof of implicit differentiation or is it simply an application of the chain rule? If it's the former, could you give or point me to the proof? If the latter, could you explain exactly how conceptually it works (or point to a link that does so)?

I stumbled upon this calculus implicit differential question: find $\frac{du}{dy}$ of the function $u=\sin <!--\; \; -->(y^2+u)$.
The answer is $\frac{2y\cos <!--\; \; -->(y^2+u)}{1-\cos <!--\; \; -->(y^2+u)}$. I understand how to get the answer for the numerator, but how do we get the denominator part?
And can anyone point out the real intuitive difference between chain rule and implicit differentiation? I can't seem to get my head around them and when or where should I use them.

Consider the curve given by the equation xy - 3x + 2y = -2. Find the equation of the tangent in the point (2,1).
So through implicit differentiation I've found that (if I'm correct) y' = (1-y)/(x-2)
How do I proceed to find the tangent line since there are still 2 variables in the function?

$Y^3+XY+X^2=0$
and I was suppose to find dY/dX and substitute the value 10 for X after differentiating. Weirdly enough, this equation does not let me get rid of the variable Y after solving and hence I couldn't find a valid answer.

Would request you to help me approach the problem. Also, I would like to do so without finding the roots of Y and solving, as someone pointed out earlier in class.

Show that $\sqrt{1+x}<1+\frac{x}{2}$ if $x>0$

I am doing implicit differentiation and I want to know the fastest way to simplify this and find y'. My online algebra calculator fails to ever solve problems the easy way. Hoping a math pro on here could show me "the easy way".
$\frac{x+y+y^{\prime }}{xy}=e^{7x-<!--\; -\; -->y}(7-<!--\; -\; -->y^{\prime })$
$=$
$\frac{1}{x}+\frac{y^{\prime }}{y}=7e^{7x-<!--\; -\; -->y}-<!--\; -\; -->y^{\prime }{e}^{7x-<!--\; -\; -->y}$
$y^{\prime }=?$