Proof that the set is not a manifold How to show that set:
$M = \left\{(x,y,z): x^2+y^2+3z^3 = xy + 6z^{\frac{1}{3}}, z \neq 0\right\} \cup \left\{(0,0,0)\right\}$ is not a manifold?
I know the problem is with point $(0,0,0)$. I think i can even show that you cannot define $x$ as $y$ and $z$ function, I mean $x=x(y,z)$. And the same with $y=y(x,z)$. But how to show that you cannot define $z=z(x,y)$?
Thank you for help.

I think that $M$ IS a manifold.
The map $z \mapsto z^3$ is a bijection from $\mathbb{R}$ to itself. Hence, doing the change of variables $z' = z^{1/3}$, we get another equation for $M$, that is:
$$ M = \{(x,y,z'): x^2 + y^2 + 3z'^9 = xy + 6z', z' \neq 0\} \cup \{(0,0,0)\}.$$
But we have in fact the equality
$$ M = \{(x,y,z') : x^2 + y^2 + 3z'^9 = xy + 6z'\},$$
since the only real solution of $x^2 + y^2 = xy$ is the couple $(0,0)$.
With this equation, we see that $M$ is a manifold since the map $f(x,y,z') = x^2 + y^2 + 3z'^9 - xy - 6z'$ is a submersion at all points of $M$.
Edit: As remarked in a comment, this only shows that $M$ is a topological submanifold of $\mathbb{R}^3$. But I think that $M$ is nevertheless a smooth manifold.
Denote by $M'$ the set $\{(x,y,z) : x^2 + y^2 + 3z^9 = xy + 6z\}$ (this was written $M$ before but now I want to make explicit the homeomorphism between $M$ and $M'$). Since $\partial_z f$ is not zero at $(0,0,0)$, there is a smooth function $g: U \rightarrow V$, with $U$ an open set in $\mathbb{R}^2$ containing $(0,0)$, $V$ is an open set in $\mathbb{R}$ containing $0$ and an open set $W$ in $\mathbb{R}^3$ such that $M' \cap W = \{(x,y,g(x,y)): (x,y) \in U\}$.
Let $\psi: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be the smooth homeomorphism $(x,y,z) \mapsto (x,y,z^3)$. One has $M = \psi(M')$ and $M \cap \psi(W) = \{(x,y,g^3(x,y)): (x,y) \in U\}$.
Edit 2: This is much more easier. Let $g(x,y,z) = (x^2 + y^2 + 3z^3 - xy)^3 - 216z$. Then $M$ is $g^{-1}(0)$ and $\partial_z g$ does not vanish at $(0,0,0)$.
