Surfaces by implicit equations

We have proved that the inverse of a regular value of a function $f:O\subset{\mathbb R}^3\rightarrow{\mathbb R}$ is a surface. Consider the prototype of function $f(x,y,z)=x^n+y^n+z^n$ where $n\in {\mathbb N}$ and let $S_a=f^{-1}(\{a\})$. The gradient of $f$ is $$\nabla f(x,y,z)=n(x^{n-1},y^{n-1},z^{n-1}).$$

First we compute the critical points, that is,  $\nabla f(x,y,z)=(0,0,0)$, obtaining  that there do no exist if $n=1$ and for $n>1$, the only critical point is $(0,0,0)$. Then 

1. For $n=1$, any value $a\in {\mathbb R}$ is regular. Here $S_a$ is a plane.
2. For $n>1$, any value $a\not=0$ is regular and $S_a\not=\emptyset$ if $a>0$. Of course, for $n=2$, we have the sphere, but if $n$ is large, $S_a$ seems a cube, but a smooth cube! 

Some pictures for $n=4$ and $n=10$:

[Remark. We observe that the set of regular value is 'big', in the above case, ${\mathbb R}-\{0\}$. There is a result that informs us about the size of $f(A)$, where $A\subset O$ is the set of critical points: the Sard's theorem says that $f(A)$ has Lebesgue measure $0$. For example, the compliment of the set of regular value has not interior points.]

Consider now the function $g(x,y,z)=x^n+y^n-z^n$ that gives the hyperboloid of one and two sheets. The set of regular values is ${\mathbb R}-\{0\}$ again. Using Mathematica, I put some pictures when $a=1$ and $n=4$ and $n=10$ 

and when $a=-1$ and $n=4$ and $n=10$