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Let k be a field. I want to find the quotient ring $k[x,y]/(x,y^2)$.

Here is my attempt: We have $(x) \subset (x,y^2)\subset k(x,y)$ so $k[x,y]/(x,y^2)\cong (k(x,y)/(x))/((x,y^2)/(x))$ hence I may guess:

$k[x,y]/(x,y^2)\cong k[y]/(y^2) $

but I still dont have rigorous proof for this formula.

I looking for some hints.

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    The generators of the ideal you quotient by tell you what elements are "glued" to $0$. So here, we get the relations $x = 0$ and $y^2 = 0$. But now we have an odd situation: $k[x, y]$ extends $k[y]$ with a new indeterminate $x$, but we also declare $x=0$. That is superfluous. So we would simply map $x$ to $0$ in our chosen isomorphism. – 2017-02-12

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You are correct. We apply the third isomorphism theorem: $$k[x,y]/ (x,y^2) \cong k[x,y]/(x)/(x,y^2)/(x) =k[y]/(y^2).$$

Thus, the quotient consists of just linear terms $1+ay$ for $a \in k$.

Morally, this makes sense since taking the quotient induces the map $x \mapsto 0$, which is linear so it extends linearly: $p[x] \mapsto 0$ for all polynomials $p$ that vanish at the origin (without constant terms.) On the other hand, the same goes for all the quadratic terms in $y$.

Another way to see it is that we want to simultaneously solve $x=0$ and $y^2=0$.