Regular Value Theorem application I'm getting ready for my Differential Geometry exam and after trying to carry out the exercises from last year's final exam, I have come up with several questions.
In one of the questions of the exam, we are given $f:\mathbb{R}^3\rightarrow\mathbb{R}$, defined as $f(x,y,z)=xyz+x^3+y^3+z^3$, and for each $\alpha\in\mathbb{R}$, we denote $F_\alpha=f^{-1}(\alpha)$. The first section follows from the regular
 value theorem, but I'm not sure that the second section does too. It asks us to show that $F_0\setminus\{(0,0,0)\}$ is a smooth submanifold of $\mathbb{R}^3$. The reason why I say that
 I am not sure that this second section is that easy is because the regular value theorem only works when you're trying to show $f^{-1}(\alpha)$ is submanifold, but what happens when you take that "problematic" point out?
For the third section of question 2, we are asked to show that for each $\alpha$ and $\beta$ in $\mathbb{R}\setminus \{0\}$, $F_\alpha$ and $F_\beta$ are difeomorphic. I think this is an easy question but I don't know how to show it.
For the next sections prior to the theoric part of the exam, I have to find the tangent plane to $F_0\setminus\{(0,0,0)\}$ in each point of the submanifold, and also that if $Y$ is a vector field of $\mathbb{R}^3\setminus\{(0,0,0)\}$ such that it is
 tangent to $F_\alpha$ for each $\alpha\in\mathbb{R}$, then $[X,Y]$ is also tangent to $F_\alpha$ for each $\alpha\in\mathbb{R}$, where $X=x\frac{\partial}{\partial  x}+y\frac{\partial}{\partial  y}+z\frac{\partial}{\partial  z}$, which is tangent to $F_\alpha$. How can I show these last two sections? I have been trying to do it in a way that is probably too calculistic (maybe it is the 
 right way, but I am not sure).  
 A: Set $M = \mathbb{R}^3 \setminus \{ (0,0,0) \}$ and let $g = f|_{M}$. Note that $M$ is an open subset, hence a submanifold and for each $0 \neq \alpha$ we have $F_{\alpha} = f^{-1}(\alpha) = g^{-1}(\alpha)$ while $f^{-1}(0) \setminus \{ (0,0,0) \} = F_0 \setminus \{ (0,0,0) \} = g^{-1}(0)$. We have
$$ dg|_{(x,y,z)} = (yz + 3x^2, xz + 3y^2, xy + 3z^2). $$
If $dg|_{(x,y,z)} = (0,0,0)$, we must have
$$ yz + 3x^2 = xz + 3y^2 = xy + 3z^2 = 0. $$
Hence, $yz, xz, xy$ must be non-positive. A case-by-case analysis shows that this is possible if and only if $x = y = z = 0$ and so $g$ doesn't have any critical points and each $g^{-1}(\alpha)$ is a two-dimensional submanifold of $M \subseteq \mathbb{R}^3$. In particular, each $F_{\alpha}$ (for $\alpha \neq 0$) and $F_{0} \setminus \{ (0,0,0) \}$ is a two-dimensional manifold.
To show that $F_{1}$ is diffeomorphic to $F_{\alpha}$ for $\alpha \neq 0$, note that $f$ is a homogeneous polynomial of degree $3$ so $f(\lambda(x,y,z)) = \lambda^3 f(x,y,z)$ for all $(x,y,z) \in \mathbb{R}^3$ and $\lambda \in \mathbb{R}$. Let $\omega = \alpha^{\frac{1}{3}}$. Then the map $(x,y,z) \mapsto \omega (x,y,z)$ is a diffeomorphic of $\mathbb{R}^3$ that maps the level set $F_1$ to $F_{\alpha}$. 
The tangent plane of $F_{0} \setminus \{ (0, 0, 0) \}$ at $(x,y,z) \in F_{0} \setminus \{ (0,0,0 \}$ is  $\ker dg|_{(x,y,z)} $. 
