Is this surface diffeomorphic to a 2-sphere? 
Let  $f:\mathbb{R}^3\to \mathbb{R}$ be defined by $f(x,y,z)=x^4+y^6+z^8$. Let $M=f^{−1}(1)$.
Is $M$ is diffeomorphic to a sphere $S^2$?

I tried to solve this problem, but I realized that I have no tools to solve it.
The constant rank theorem tells me $M$ is a smooth 2-dimensional manifold, but does not tell me how it looks like.
And more generally, when is $N=\{x,y,z\in\mathbb{R}^3\mid ax^n+by^m+cz^l=1\}$ diffeomorphic to a sphere? What tools can I use to solve this problem?
Thank you for reading. Hoping get some shedding light in your reply.
 A: Morse Theory is a very suitable tool to solve this kind of problems. We shall find a Morse function on $M$ and we can compute the indices of its critical points. Take
$$
\begin{array}{rccc}
h:&M&\longrightarrow&\mathbb{R}\\
  &(x,y,z)&\longmapsto & x
\end{array}.
$$
Note that $T_{(x,y,z)}M=\ker d_{(x,y,z)}f$. Since
$$
df=4x^3dx+6y^5dy+8z^7dz
$$ 
its kernel is generated by
$$
3y^5\frac{\partial}{\partial x}-2x^3\frac{\partial}{\partial y}\ \mbox{and}\ 2z^7\frac{\partial}{\partial x}-x^3\frac{\partial}{\partial z},\ \mbox{at points with}\ x\neq 0
$$
$$
\frac{\partial}{\partial x}\ \mbox{and}\ 4z^7\frac{\partial}{\partial y}-3y^5\frac{\partial}{\partial z},\ \mbox{at points with}\ x=0
$$
Now, we have $dh=dx$ which applied to the vector fields above gives the functions $3y^5,2z^7$ at points with $x\neq 0$ and the constants $1,0$ at points with $x=0$. From this it follows that $(x,y,z)\in M$ is a critical point of $h$ if and only if $y=z=0$. So $h$ has exactly two critical points, which are $(1,0,0)$ and $(-1,0,0)$.
Finally, since $M$ is compact --I assume you know how to prove this-- and admits a Morse function with only two critical points, it must be a sphere (in this very simple case there's no need to compute the indices of the critical points, although they are 0 and 2, of course).
As for the general case $ax^n+by^m+cz^l=1$ you can carry out a similar computation and decide what surface you get in terms of critical points and its indices. However you'll need to be careful with compactness (what happens if one of the exponents is odd?) and probably you'd like to assume $abc\neq 0$.
A: I'll recast the problem in more general terms. We are given a compact submanifold $M\subset \mathbb R^n\setminus \{0\}$ which intersects every ray $\{t x:t>0\}$, $x\in\mathbb R^n\setminus\{0\}$, exactly once, and transversely. The claim is that $M$ is diffeomorphic to $\mathbb S^{n-1}$.
Transverse intersection of submanifolds $M_1,M_2$ means that at every point $p\in M_1\cap M_2$ the union of  tangent spaces $T_pM_1$ and $T_pM_2$ spans the tangent space of the ambient manifold ($\mathbb R^n$ for us). In our situation this requirement amounts to $T_pM$ not containing the vector pointing from $p$ to the origin. And if $M$ is defined by equation $f=c$, this can be rephrased again by saying that $\nabla f(x)$ is never orthogonal to $x$; the latter is easy to check in your example.
Consider the radial map $G( x)=\dfrac{ x}{| x|}$  which radially projects $\mathbb R^n\setminus\{0\}$ onto the sphere $\mathbb S^{n-1}$.  This is a submersion: a smooth surjective map such that the rank of differential is equal to dimension of the target space. Indeed, the derivative matrix of $G$ is $\dfrac{|x|^2I-x\otimes x}{|x|^3}$, which has one-dimensional kernel: namely, the vectors collinear to $x$. 
Let $g=G_{|M}$, the restriction of $G$ to $M$. The differential also restricts, and by the transversality condition the differential of $g$ has trivial kernel. Since $g:M\to\mathbb S^{n-1}$ is a bijection by assumption, and both spaces are compact, we conclude that $g$ is a homeomorphism. Having invertible derivative at every point, it is also a diffeomorphism. 

The above applies to $ax^n+by^m+cz^l=1$ whenever this surface is compact. When it is not compact, it can't be homeomorphic to the sphere, let alone diffeomorphic to it.

Added remark: I think it suffices to assume that $M$ is a compact submanifold whose intersection with every ray $\{t x:t>0\}$, $x\in\mathbb R^n\setminus\{0\}$ is transverse when it is nonempty.  Indeed, this assumption implies, via the above argument, that $g:M\to\mathbb S^{n-1}$ is an open map. Hence $g(M)$ is  open in $\mathbb S^{n-1}$, but being also compact, it must coincide with $\mathbb S^{n-1}$. It remains to show that $g$ is injective, but I'm drawing a blank here.
