Let $I = \langle y-x^2 \rangle \subset \mathbb{k}[x, y]$, where $\mathbb{k}$ is a field.
Denote $A= \mathbb{k}[x, y] / I$.
I need to show $A \simeq \mathbb{k}[\bar x] $, where $\bar x$ is the equivalence class of $x$ in $A$.
I wonder if the following proof is correct.
I define $\phi: \mathbb{k}[x, y] \to \mathbb{k}[\bar x] $ by $\phi(p) = p(\bar x, \bar x^2)$. I try to show $\ker \phi = I$.
Assume $\phi(p) = 0$. Now we can write $p = q*(y-x^2) + r$, where $r \in \mathbb{k}[x]$ (This is where I'm uncertain: can we always drop the degree of $y$ and thus eliminate it? If we use the lexicographic ordering in the multivariable division theorem or what ever Gröbner basis it is called.) Anyhow then we can say that $r=0$, because $p(\bar x, \bar x^2) = 0$. And thus $p \in I$. The other way inclusion $I \subset \ker \phi$ is clear and by the fundamental theorem of ring homomorphisms the proof is complete.
Here is a possible alternative. Consider the maps of rings, determined by the image of the generators:
$f:\Bbb k[x,y]/(y-x^2)\to \Bbb k[X]$, $f(x) = X$, $f(y)=X^2$, which is well defined, since $f(y-x^2)=f(y)-f(x)^2=(X^2)-(X)^2=0$,
$g:\Bbb k[X]\to \Bbb k[x,y]/(y-x^2)$, $f(X) = x$.
We compute than the compositions on the generators:
$(gf)(x)=g(f(x))=g(X)=x$, $(gf)(y)=g(f(y))=g(X^2)=x^2=y$, so $gf$ is the identity,
$(fg)(X)=f(g(X))=f(x)=X$, so $fg$ is the identity.