This question is motivated by exercise 1.1 in Hartshorne Algebraic Geometry. One has to prove that $k[x,y]/(y-x^2)$ is isomorphic to a polynomial ring in one variable. I know that this can be done by viewing $k[x,y]/(y-x^2)$ as the image under the surjection $k[x][y] \to k[x][y]/(y-x^2)$ given by $x \mapsto x$, $y \mapsto x^2$ (that is, $f(x,y) \mapsto f(x,x^2)$). Since $x^2 \in k[x]$, the image under this evaluation map is just $k[x]$.
My question is, why does it apparently not work to view $k[x,y]/(y-x^2)$ as $k[y][x]/(y-x^2)$, the image under the map $k[y][x] \to k[y][x]/(y-x^2)$ given by $x \mapsto \sqrt{y}$, $y \mapsto y$? Since $\sqrt{y} \notin k[y]$, I think the evaluation can't produce a ring isomorphic to $k[y]$. But it seems to me that if $k[x,y]/(y-x^2)$ is isomorphic to $k[x]$, then it should be isomorphic to $k[y]$ also, and therefore cannot be isomorphic to the ring obtained by this evaluation.