I'm approaching localization of rings using multiplicatively closed sets, and the obvious case of when we take the complementary of a prime ideal of a ring, that is always multiplicatively closed.
The question at hand asks to prove that the ring $k[x,y]/(xy^2,x^2y)$ localized at $(x)$ is (or isn't) isomorphic to the ring $k[x]$ localized at $(x)$.
I think the second ring could be characterized as the ring of one-variable polynomials where the denominator, expressed as sum of monomials, needs to have a non zero constant, and the numerator can be whatever. I have trouble visualizing the first one. The quotient ring seems to describe the ring of polynomials in two variables, $x,y$, whose mixed terms can only be of degree one (So, just $xy$ and not higher powers). Localizing it to $(x)$ means in this case... What? That there can't be polynomials at denominator with all its monomials multiplying x? There needs to be at least an y, or a constant in there?
Notice that $k[x,y]/(xy^2,x^2y)$ has the zero divisor $y^2$ since it becomes zero after multiplying it with the non-zero element $x$. In particular it is also a zero divisor after inverting at the prime ideal $(x)$. Since $k[x]$ is a domain it can not be isomorphic to $(k[x,y]/(xy^2,x^2y))_{(x)}$.
$\textbf{Edit:}$ I think the ring collapses since $y^2\notin(x)$ and therefore will be inverted.