Are there only two irreducible affine singular cubic curves? 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?
 A: 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 $x^3+y^3=3axy$ for any $a \neq 0$. It has the singular nodal point $(0,0)$ and again 3 points at infinity.
A: Take an affine irreducible singular cubic curve  $C$ of degree $3$ in $\mathbb{A}^2$, and embedd $\mathbb{A}^2$ into $\mathbb{P}^2$. The closure is an irreducible curve $\Gamma\subset \mathbb{P}^2$ of degree $3$ with one singular point in $\mathbb{A}^2$, so no singular point on the line $L=\mathbb{P}^2\setminus\mathbb{A}^2$ (by Bézout with the line through the points or by genus formula, a cubic has only one singular point). It is moreover one of your two examples. So the curve $C$ is equal to $\Gamma\setminus L$ for some line $L$ in $\mathbb{P}^2$. If you want to classify $C$ up to affine change of coordinates, you classifiy the pairs of cubics $\Gamma$ and lines $L$ up to change of coordinates. As the group of automorphisms of $\mathbb{P}^2$ preserving a cuspidal curve is of dimension $1$, and the set of lines if of dimension $2$, you get a family of dimension $1$ in this case. For the nodal case, there are only finitely many automorphisms of $\mathbb{P}^2$ preserving the curve, so you get a family of dimension $2$ of examples.
You can then ask for a classification up to isomorphisms. In this case, the isomorphisms between the affine curves are the same as the isomorphisms between the singular curves which map the points at infinity on the points at infinity. This arleady makes a difference between the cases with $3$, $2$ or $1$ point at infinity, but moreover you again can check that the group of automorphisms of the cuspidal cubic is of dimension $2$ (corresponds to the automorphisms of $\mathbb{A}^1$ when removing the point) and of the nodal cubic is $1$ (automorphisms of the complement of the singular point: a torus and an involution). In the first case, you get only finitely many possibilities, but in the second case, you still get infinitely many pairwise non-isomorphic affine nodal cubic curves.
