It's known that every irreducible projective singular cubic curve in $\mathbb{CP}^2$ is isomorphic (actually projectively equivalent) to either a cuspidal cubic $$y^2z=x^3,$$ or a nodal cubic $$y^2z=x^3+x^2z.$$ If we focus on the affine case, is it still true that every irreducible affine singular cubic curve in $\mathbb{C}^2$ is isomorphic to either $$y^2 = x^3$$ or $$y^2=x^3+x^2?$$ If not, what is a counter-example?



Are there only two irreducible affine singular cubic curves?
If you remove some (non-singular) points from the curve $y^2=x^3$, you get a singular cubic which is still affine (it is an open affine subset), yet it is not isomorphic to the original curve. Indeed, any isomorphism would induce an isomorphism between the regular loci of the curves. These loci are just projective lines minus finitely many points, and the isomorphism class determines the number of missing points.


**EDIT.** This argument is partly relevant because I forgot that the OP asks for affine *plane* curves. However, one can give a counter-example to the OP's question.


Note that the cubics $y^2=x^3$ and $y^2=x^3+x^2$ have only one point at infinity. Let's try for a singular plane cubic with three points at infinity. Consider the family of plane cubics
$$C\_k : x^3+y^3+kxy+1=0.$$
The singular fibers are given by $k \in \{-3,-3\zeta\_3, -3\overline{\zeta\_3}\}$. For example, when $k=-3$ the (unique) singular point is $(x,y)=(1,1)$ and is a node. The regular locus of $C\_{-3}$ is isomorphic to the projective line minus 5 points, hence $C\_{-3}$ cannot be isomorphic to the above cubics. I don't know however if $C\_{-3}$ (or other singular plane cubic curves) can be isomorphic to open subsets of these two cubics.


**EDIT 2.** A simpler example is given by [Descartes's folium](https://en.wikipedia.org/wiki/Folium_of_Descartes) $x^3+y^3=3axy$ for any $a \neq 0$. It has the singular nodal point $(0,0)$ and again 3 points at infinity.


