Is $Z(x^2-y^3)$ isomorphic to $Z(y^2-x^3-x^2)$ over the complex numbers?

I'm having trouble determining if the algebraic sets $Z(x^2-y^3)\subset \mathbb{A}^2$ and $Z(y^2-x^3-x^2)\subset\mathbb{A}^2$ are isomorphic over $\mathbb{C}$. My guess is that this boils down to determining if $\mathbb{C}[x,y]/(x^2-y^3)$ is isomorphic to $\mathbb{C}[x,y]/(y^2-x^3-x^2)$ but then again, I'm stuck.

-
Isomorphic as sets? as abelian groups? as rings? – Gerry Myerson Oct 24 '12 at 3:05
Isomorphic as affine varieties. – Victor Oct 24 '12 at 3:10
This is a dublicate of my question math.stackexchange.com/questions/128918/…. There I also give an alternative proof. – Martin Brandenburg Oct 24 '12 at 8:02
Thanks for pointing it out! – Victor Oct 24 '12 at 8:08

Thinking geometrically, we expect these varieties are not isomorphic, due to the fact that the first is a cusp, while the second is a node. One way to verify this is to consider the tangent cone of each. In the first case, we get $TC_{(0,0)}=V(x^2)$ which is interpreted as the line $x=0$ with multiplicity $2.$ In the second case, we get $TC_{(0,0)}=V(y^2-x^2)=V((y-x)(y+x))$ which is two distinct lines.
@Victor, actually, you should ignore my previous remark; we cannot use the tangent space instead, since both defining polynomials have no linear terms. Thus both curves have tangent space $\cong\Bbb A^2,$ so the tangent cone is quite convenient here. – Andrew Oct 24 '12 at 3:56