I'm working through the derivation of the fact that $$\nabla f$$ is perpendicular to the level surfaces of f.
One of the steps says that "in general $$\nabla f \neq0$$
where $$\nabla f$$ is the gradient of f."
Why is this true?
Is it because we can't have a saddle/max/min on a level surface?
On
$\Vert x \Vert = 1$ is a level set for $f(x) = (\Vert x \Vert - 1)^2$. All points in that set are minima of the function, and the gradient is zero everywhere on that set. So you cannot say that $\nabla f \neq0$ “in general on a level set.”
On the other hand, the null vector is perpendicular to every other vector, so that these points can be ignored when proving that the gradient is perpendicular to a level surface.
When some function $f:\>{\mathbb R}^3\to{\mathbb R}$ is given you have level sets $$N_c:=\bigl\{(x,y,z)\bigm| f(x,y,z)=c\bigr\}\qquad(c\in{\mathbb R})\ .$$ If $p$ is an arbitrary point in the domain of $f$ then $p\in N_c$ for the value $c:=f(p)$. If it is true that $\nabla f(p)\ne0$ then the implicit function theorem guarantees that $p$ is the center of a rectangular box $R$ such that $N_c\cap R$, i.e., the part of $N_c$ contained in $R$, is a smooth surface.
The points $q$ in the domain of $f$ where $\nabla f(q)=0$ are called critical points of $f$. These critical points are solutions of the $3\times3$ equation system $$f_x(x,y,z)=0,\quad f_y(x,y,z)=0, \quad f_z(x,y,z)=0\ .$$ For most functions they are isolated, but it is easy to cook up examples where this is not the case (see Martin R's answer). If $q$ is a critical point of $f$ then $N_{f(q)}$ need not be a fine surface in the neighborhood of $q$. Consider the following example: $$f(x,y,z):=z^2-x^2-y^2,\qquad\nabla f(x,y,z)=(-2x,-2y,2z)\ .$$ This $f$ has the single critical point $q=(0,0,0)$. The level set $N_0$ is in fact a double cone emanating from $q$, hence has a singularity at $q$. All other level sets are hyperboloids and smooth surfaces throughout.