Gradient vector and regular surfaces I'm looking through my notes and I don't fully understand what this is telling me. What is the purpose of the gradient vector here and why is $S^2$ regular because of the following? I'm looking for the big picture and any underlying propositions or theorems that might be related. 

Let $S^2$ = {$(x, y, z) \in \Bbb{R}^3$ | $x^2 + y^2 + z^2 = 1$} be the unit sphere. Then we can write $S^2$ = {$(x, y, z) \in \Bbb{R}^3$ | $f(x,y,z) = 0$} where $f$ : $\Bbb{R}^3 \to \Bbb{R}$ : $f(x,y,z) = (x^2 + y^2 + z^2-1)^2$.
Then...
$\triangledown f(x,y,z) = (\frac{∂f}{∂x}, \frac{∂f}{∂y}, \frac{∂f}{∂z})$ = 2($x^2 + y^2 + z^2 - 1$)($2x$, $2y$, $2z$)
$= (0)(2x, 2y, 2z)$ for all $(x, y, z) \in S^2$
$= (0, 0, 0)$ for all $(x, y, z) \in S^2$
$S^2$ = {$(x, y, z) \in \Bbb{R}^3$ | $g(x, y, z) = 0$} where $g(x, y, z)$ = $x^2 + y^2 + z^2 - 1$. 
$\triangledown g(x, y, z) = (2x, 2y, 2z) = (0, 0, 0)$ if and only if $(x, y, z) = (0, 0, 0)$ but $(0, 0, 0) \notin S^2$. Therefore, $S^2$ is regular by our proposition. 

The relevant theorem is the level set theorem which in this special case will say if  $f: \mathbb{R}^3 \to \mathbb{R}$ is a smooth map and for all $p \in f^{-1}(c)$ we have $Df(p) \not = \textbf{0}$ then $f^{-1}(c)$ is a regular surface. 
