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Mathematically this can't be counted as manifold because $\textbf{rank}D_f(0,0) = (3\cdot 0^2\quad 3\cdot0^2) = 0$.

However, I see no problem at this point the reason being the manifold is a straight line $(y = -x)$

Leon
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    What is $f$? And why “Mathematically”? Are you suggesting that, say, from the point of view of Physics it is a manifold? – José Carlos Santos Feb 22 '21 at 10:24
  • You probably know a criterion for a level curve to be a manifold? – Pedro Feb 22 '21 at 10:26
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    $M$ is a manifold (indeed, this is a straight line). However, you are right, the map $(x,y) \mapsto x^3 + y^{3}$ is not a submersion at $0$. The thing is that, if you can write $M = { (x,y) \in \mathbb{R}^2 : \psi(x,y) = 0}$, then if $0$ is a regular value of $\psi$ it follows that $M$ is a submanifold of $\mathbb{R}^2$. However, it is not true that if $M$ is a submanifold of $\mathbb{R}^2$ then $0$ must be a regular value of $\psi$ (you have a counter-example). –  Feb 22 '21 at 10:27

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