How can I show that M\{(0,0)} is a submanifold?

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I have the following problem:

Let $$M=\{(x,y):y^2=x^3\}$$ Show that $M\setminus \{(0,0)\}$ is a submanifold.

I wanted to do this as follows: Let us denote the function $$F:U\rightarrow \mathbb{R};\,\,\,(x,y)\mapsto y^2-x^3$$ Where $U=\mathbb{R}\setminus \{(0,0)\}$. Then $F^{-1}(0)=M$ and furthermore $$DF=(-3x^2,\,2y)$$We remakr that $(-3x^2,\,2y)$ has rank 1 iff $x\neq 0$ or $y\neq 0$ but since $(0,0)\notin U$ we get that F is a subersion and thus M is a submanifold of dimension 1.

Is this correct like this?

Thanks a lot, and sorry for that much questions but I had a big struggle to understand submanifolds and now I think that it get clearer and clearer (so only if this is correct what I have done).

Additional:

We have the following definition: Def: $M\subset \mathbb{R}^n$ is submanifold of dimension k if for all $a\in M$ one of the following three equivalent points holds:

  1. For each $a\in M$ there exists an open neighbourhood $U\subset \mathbb{R}^n$ and a diffeomorphismen $\phi:U\rightarrow V$ where V is open in $\mathbb{R}^n$ such that $$\phi(U\cap M)=V\cap(\mathbb{R}^k\times \{0\})$$
  2. For each $a\in M$ there exists an open Neighbourhood $U\subset \mathbb{R}^n$ and a submersion $F:U\rightarrow \mathbb{r}^{n-k}$ at a such that $$U\cap M=U\cap\{F=0\}$$
  3. For each $a\in M$ there exists an open $O\subset M$ and and open $O'\subset \mathbb{R}^k$ and a homeomorphism $\phi:O'\rightarrow O$ such that $\phi$ is a immersion at $\phi^{-1}(a)$ and $O=\phi^{-1}(O)$