We have the filtration: $$k[x]_{(x)} \supset (x)_{(x)} \supset (x^2)_{(x)} \supset (x^3)_{(x)} \supset \cdots $$ and so we have $${\rm gr}_{(x)_{(x)}} k[x]_{(x)} = \frac{k[x]_{(x)}}{(x)_{(x)}} \oplus \frac{(x)_{(x)}}{(x^2)_{(x)}} \oplus \frac{(x^2)_{(x)}}{(x^3)_{(x)}} \oplus \cdots$$ which simplifies to $${\rm gr}_{(x)_{(x)}} k[x]_{(x)} = \left(\frac{k[x]}{(x)}\right)_{\frac{(x)}{(x)}} \oplus \left(\frac{(x)}{(x^2)}\right)_{\frac{(x)}{(x^2)}} \oplus \left(\frac{(x^2)}{(x^3)}\right)_{\frac{(x^2)}{(x^3)}} \oplus \cdots.$$
I know $\left(\frac{k[x]}{(x)}\right)_{\frac{(x)}{(x)}}$ simplifies to just $k$.
However, what does $\left(\frac{(x)}{(x^2)}\right)_{\frac{(x)}{(x^2)}}$ simplify to?
Am I even thinking about this the right way?
I'm not sure this is right, unless I'm misunderstanding some notation. What does it mean to localize with respect to $(x)/(x^2)$? Which ring is that meant to be a multiplicative subset or prime ideal of? What is true is that $(A/B)_\mathfrak{p} = A_\mathfrak{p}/B_\mathfrak{p}$, since localization is exact. So I think it would be more correct to say
Now, each $(x)^n/(x)^{n+1}$ is a $1$-dimensional $k$-vector space (spanned by $x^n$) where $x$ acts trivially – therefore, as a $k$-vector space, we have
$${\rm gr}_{(x)_{(x)}} k[x]_{(x)} \cong \bigoplus_\mathbb{N} k.$$
In order to really determine the associated graded ring though, we must also understand the multiplicative structure. So, to check your understanding: what is the product operation $(\bigoplus_\mathbb{N} k) \times (\bigoplus_\mathbb{N} k) \to \bigoplus_\mathbb{N} k$? And what well-known ring is ${\rm gr}_{(x)_{(x)}} k[x]_{(x)}$ isomorphic to?