Is k[[x]] ever a finitely generated k[x](x) module? For k a field, the localization k[x]_{(x)

Explanation:
In fact, k[[x]] is never even countably generated as a ${\displaystyle k{\left[x\right]}_{\left(x\right)}}$-module.
To prove this, suppose k[[x]] were generated by countably many elements ${\displaystyle {\left\{c_i\right\}}_{i\in \mathbb{N}}}$ as a ${\displaystyle k{\left[x\right]}_{\left(x\right)}}$-module, and let ${\displaystyle k_0\subseteq k}$ be the subfield generated by all the coefficients of the ${\displaystyle c_i}$. For any ${\displaystyle f\in k\left[\left[x\right]\right]}$, then, there exist ${\displaystyle a,b_1,\cdots ,b_n\in k\left[x\right]}$ with ${\displaystyle a\ne 0}$ such that
${\displaystyle af=\sum _{i=1}^n{b}_i{c}_i}$
(here a is the common denominator of the coefficients when you write f as a ${\displaystyle k{\left[x\right]}_{\left(x\right)}}$-linear combination of the ${\displaystyle c_i}$). If ${\displaystyle f\in k_0\left[\left[x\right]\right]}$, we can consider this equation as a system of linear equations with coefficients in ${\displaystyle k_0}$ (coming from the coefficients of f and the ${\displaystyle c_i}$) in the (finitely many) coefficients of a and the ${\displaystyle b_i}$. A basis for the solutions of such a system can be computed by Gaussian elimination, and this process is unchanged by extending the base field. In particular, since there exists a solution to this system over k where one of the coefficients of a is nonzero, such a solution also exists over k0. This just means that f is also a ${\displaystyle k_0{\left[x\right]}_{\left(x\right)}}$-linear combination of the ${\displaystyle c_i}$.
Thus ${\displaystyle k_0\left[\left[x\right]\right]}$ is also countably generated as a ${\displaystyle k_0{\left[x\right]}_{\left(x\right)}}$-module (by the same countably many elements ${\displaystyle c_i}$). But ${\displaystyle k_0}$ is countable, so ${\displaystyle k_0{\left[x\right]}_{\left(x\right)}}$ is countable and thus so is any countably generated ${\displaystyle k_0{\left[x\right]}_{\left(x\right)}}$-module. Since ${\displaystyle k_0\left[\left[x\right]\right]}$ is uncountable, this is a contradiction.