Determine whether two rings are isomorphic $k$ is a field, char$(k)\neq 2$.
Can $k[x,y]/(y-x^2)$ be isomorphic to  $k[x,y]/(x^2+y^2-1)$ by some ring homomorphism?
Edit: Since jspecter gives a nice and quick proof, 
I will ask another question 

Can $k[x,y]/(y^2-x^2(x+1))$ be isomorphic to $k[x,y]/(x^2+y^2-1)$?

MY motivation is that  two picture defined $y-x^2$ and $x^2+y^2-1$ are not "homeomorphic" as we view in $\mathbb{R}^2$, maybe it will induce that the coordinate rings are not isomorphic?
 A: They are not isomorphic as $k$-algebras, at least.  (From a geometric perspective, this is the more natural question....)
The first ring -- say $R$ -- is isomorphic to $k[t]$, so is the ring of regular functions on the affine line $\mathbb{A}^1$ over $k$.  In other words, it corresponds to the projective line minus a single point at infinity.
The second ring -- say $S$ -- is the ring of polynomial functions on the unit circle.  Geometrically, it is the projective line minus two points at infinity.  If $\sqrt{-1} \in k$, these points are both rational and $S \cong k[t,t^{-1}]$.  Thus the unit group of $S$ is larger than just $k^{\times}$ and so they cannot be isomorphic as $k$-algebras.
In case $\sqrt{-1}$ does not lie in $k$, if they were isomorphic as $k$-algebras then by tensoring with $k(\sqrt{-1})$ they would also be isomorphic as $k(\sqrt{-1})$-algebras and one can reduce to the previous case.  But actually in this case one can do better: the Picard group of $S$ has order $2$, whereas the Picard group of $R$ is trivial, so $R$ and $S$ are not isomorphic even as rings.  (See e.g. page 6 of this paper for a treatment of the Picard group of $S$.)
This answer leaves open the possibility that $R \cong S$ as rings when $\sqrt{-1} \in k$.  Probably someone else will address this...
Added: The third ring $T = k[x,y]/(y^2-x^2(x+1))$ is the coordinate ring of an affine curve which is singular at the origin, so is not even a Dedekind domain, unlike $R$ and $S$.
