If $0$ is the unique critical value of $f$, than, with $ab>0$, $f^{-1}(a)$ and $f^{-1}(b)$ are diffeomorphics I've have a huge problem to solve (at least to me): Let $f(x,y,z) = x^3 + y^2 -zy^2$. In fact, $0$ is the only critical value of $f$. Here comes the real deal: How can I prove that, if $ab>0$, than $f^{-1}(a)$ and $f^{-1}(b)$ are diffeomorphics? To solve it, one can use the Inverse Function Theorem or the Implicit Function Theorem.
After that, one more question: How to write the equation of the tangent plane to $X = f^{-1}(a)$ in a point $p=\left(\alpha,\beta,\gamma \right)$?
Thanks in advance!
 A: This is basically the  First Fundamental Theorem of Morse theory. As in the comments, define
$X_x=\begin{cases} \frac{\nabla f(x)}{\|\nabla f(x)\|^2} \quad x\in f^{-1}((a-\epsilon,b+\epsilon))\\ 0\quad \quad \quad \text{otherwise}\end{cases}$
This is well-defined because there are no singularities of $f$ between $a$ and $b$ so there is some $\epsilon>0$ neighbrhood on which this is still true. (why?).
All this is is the normalized gradient vector restricted to a certain piece of the domain of $f$.
Notice that $\phi:\mathbb R\times \mathbb R^3\to \mathbb R^3$ defined by $(t,x)\mapsto \frac{1}{\|\nabla f(x)\|^2}(t\partial_1f(x),\cdots,t\partial_3f(x) )$ satisfies
$X_{\phi(t,x)}=\frac{d(\phi(t,x))}{dt}$ and is a diffeomorphism for each $t\in \mathbb R.$ Note also that $\phi_{-t}\circ \phi_t=\text{id}.$
The claim is now that $\phi_{b-a}$ is a diffeomorphism onto $f^{-1}(b)$ when restricted to $f^{-1}(a).$
We have $\frac{df\circ \phi(t,x)}{dt}\overset{\text{chain rule}}{=}\langle \frac{d\phi(t,x)}{dt},\nabla f(x)\rangle=\langle X_{\phi(t,x)},\nabla f(x)\rangle.$
It follows from this that $\frac{df\circ \phi(t,x)}{dt}=\begin{cases} 1 \quad x\in f^{-1}((a-\epsilon,b+\epsilon))\\ 0\quad \quad \quad \text{otherwise}\end{cases},\ $ and so
$f\circ \phi(t,x)=\begin{cases} t+f(x) \quad x\in f^{-1}((a-\epsilon,b+\epsilon))\\ f(x)\quad \quad \quad \text{otherwise}\end{cases}.$
Now, consider the diffeomorphism $\varphi:=\phi_{b-a}|_{f^{-1}(a)}.$ If $x\in f^{-1}(a)$ then $\phi_{b-a}(b-a,x)=b-a+f(x)=b-a+a=b$ so $\varphi$ maps  $f^{-1}(a)$ into $f^{-1}(b).$
Now suppose $y\in f^{-1}(b)$ and let $x$ be such that $\phi_{a-b}(y)=x.$ Then, $\varphi(x)=\phi_{b-a}\circ \phi_{a-b}(y)=y$ so in fact, $\varphi$ maps $\textit{onto}\ f^{-1}(b)$ and we are done.
