Show that X={((x_0: x_1), (y_0:y_1)):(x_0:2+_1^2)(y_0^2+y_1^2)=x_0x_1y_0y_1)} subseteq P^1 times P^1 is a smooth curve of genus 1.

Step 1
I'm going to go ahead and show you how to do this with adjunction. The adjunction formula computes the canonical class of a hypersurface from the canonical class of the ambient variety; specifically, it says that for $D\subset <!--\; \subset \; -->X$ a divisor, we have $K_D=(K_X+D){|}_D$.
Now ${\mathbb{P}}^1\times <!--\; \times \; -->{\mathbb{P}}^1$ has Picard rank 2, so we typically specify a divisor class by an ordered pair (a,b). One way of thinking about these classes (at least in the effective case) is that a divisor of type (a,0) is the pullback of a degree a divisor (e.g. the sum of a points) from the first ${\mathbb{P}}^1$ factor, and similarly for type (0,b). These pullbacks are simply disjoint unions of lines (all in the same ruling), the fibers over the divisors on ${\mathbb{P}}^1$. Taking the union of a divisor of each type, you get a divisor of type (a,b), where now the a lines in one ruling intersect the b lines from the other ruling.
Step 2
The other thing to know is the intersection pairing: $(1,0)\cdot <!--\; \cdot \; -->(0,1)=1$ (two lines, one from each ruling, intersect in a unique point), and $(1,0)\cdot <!--\; \cdot \; -->(1,0)=(0,1)\cdot <!--\; \cdot \; -->(0,1)=0$ (two lines, both from the same ruling, never intersect). Everything else extends by linearity, so in general $(a,b)\cdot <!--\; \cdot \; -->(c,d)=ad+bc$.
To use adjunction here, we need $K_{{\mathbb{P}}^1\times <!--\; \times \; -->{\mathbb{P}}^1}$ and D. Now hopefully you know that $K_{{\mathbb{P}}^1}=-<!--\; -\; -->2$; it follows from general facts about products of varieties that $K_{{\mathbb{P}}^1\times <!--\; \times \; -->{\mathbb{P}}^1}=(-<!--\; -\; -->2,-<!--\; -\; -->2)$. Since your curve is given by an equation of bidegree (2,2), its divisor class is also (2,2). So $K_{{\mathbb{P}}^1\times <!--\; \times \; -->{\mathbb{P}}^1}+D=(-<!--\; -\; -->2,-<!--\; -\; -->2)+(2,2)=(0,0)$. To compute the degree of the restriction of this class to your curve D, we would normally now compute $(0,0)\cdot <!--\; \cdot \; -->(2,2)=0$, but this is unnecessary since the trivial class always restricts to the trivial class. Now since the only curves with trivial canonical class are genus 1, we are done.