This is from a course which closely follows Munkres' "Analysis on Manifolds". We are asked to prove / disprove whether $M = \{ (x,y) \in \mathbb{R}^2: y^2 = x^3 \} $ is a differentiable manifold, which it isn't. Theorem 24.4 from Munkres allows one to prove that for a given function $ f \colon \mathbb{R}^n \to \mathbb{R}$ of class $C^r$, $ \{ \textbf{x} \in \mathbb{R}^n: f( \textbf{x} ) = 0 \} $ is a manifold provided that $Df( \textbf{x} ) \neq \textbf{0}$ ( and the set is nonempty). Is the converse true, that if $Df( \textbf{x} ) = \textbf{0}$ somewhere on $M$ that it fails to be a manifold? Setting $g(x,y) = x^3 - y^2 $, since $Dg(0,0) = \textbf{0}$ this would provide a quick proof that $M$ is not a manifold.
I wouldn't have suspected it would be true but a solution to an older question seems to use this method. I have not found anything in Munkres about it. I have a different solution, my question is just whether the method described above is valid.
No, just take some function $f$ which defines a manifold. Then $\bar{f} = f^n$ defines the same manifold but $\bar{f}'(x) = nf(x)^{n-1}f'(x) = 0$ per $n> 1$