Prove this plane algeraic curve is not a differentiable manifold

Question

Prove the algebraic curve $\{(x,y)~|~x^2(x+1)-y^2=0\}$ in $\mathbb{R}^2$ is not a differentiable manifold.

Remark: It is evident that the given cubic curve has a singularity at $(0,0)$ which disable the existence of a tangent space at the origin, hence it is not (globally) a differentiable manifold. But how to show this by using the formal definition of a differentiable manifold? Do we need to prove that ANY "atalas" would not satisfy the "smooth transition map" criterion?

Answer 1

In a $1$-dimensional curve, no point has a connected open neighborhood which splits into four connected components when you remove a point from it.

Prove that and then use it to show your curve is not a $1$-dimensional manifold.

Answer 2

Alternative powerful criterion: A $1$-manifold in $\Bbb R^2$ must, in a neighborhood of each point, be the graph of either $y=f(x)$ or $x=g(y)$ for some smooth function $f$ or $g$. This is an exercise in applying the Inverse Function Theorem.