If scalar field $f(x,y,z)$ is differentiable then the set of point satisfying $f(x,y,z)=c$ ($c$ is constant) is smooth surface? Let $f(x,y,z), x,y,z\in \mathbb{R}$ is a scalar field. Could you prove that if $f$ differentiable then the set of points $(x,y,z)$ sastifying $f(x,y,z)=c$ ($c$ is constant) is a smooth surface?
The definition of smoothness of surface bases on parametric form of surface, but this is not in parametric form, so I'm stuck.
 A: Given a function $f:\>{\mathbb R}^3\to{\mathbb R}$ and a value $c\in{\mathbb R}$ the object
$$S:=f^{-1}\bigl(\{c\}\bigr)=\bigl\{(x,y,z)\in{\mathbb R}^3\bigm| f(x,y,z)=c\bigr\}\quad$$
is a subset of the domain ${\mathbb R}^3$ of $f$, and is called a level set of $f$. Under reasonable assumptions such level sets are smooth surfaces, but we have to expect singularities.
If $p:=(x_0,y_0,z_0)$ is a point of $S$, and if $\nabla f(p)\ne{\bf 0}$ then we have, e.g., ${\partial f\over\partial z}(p)\ne0$. In this case there is a rectangular box $R$ with center $p$ and a $C^1$-function $(x,y)\mapsto z:=\psi(x,y)$ such that $S\cap R$, i.e., the part of $S$  lying in $R$, coincides with the graph of $\psi$. This means that we have a parametric representation of $S\cap R$ as follows:
$$\quad R'\to{\mathbb R}^3,\qquad (x,y)\mapsto\bigl(x,y,\psi(x,y)\bigr)\ ,\tag{1}$$
whereby $R'$ denotes the "floor" of the box $R$. $(1)$ shows that in the neighborhood of $p$ the level set $S$ is indeed a smooth surface.
But in the critical points of $f$ there will be trouble. If, e.g., $f(x,y,z)=x^2+y^2+z^2$ then the level set belonging to $c=0$ is a single point. For  $f(x,y,z)=x^2-y^2$ and $c=0$ we obtain  the union of two intersecting planes, with all points on the $z$-axis "singular".
