While trying to understand the Langrange optimization method the following issue came to my mind.
Let's assume that the function $f:\mathbb{R}^n\to\mathbb{R}$ is continuously differentiable and $M(g)\subset\mathbb{R}^n$. $M(g)$, the constraint, is defined by $M(g):=\{x\in\mathbb{R}^n\mid g(x)=0\}$, where $g$ is also continuously differentiable and its derivative $Dg$ satisfies $Dg\neq \textbf{0}$ for all $x\in M(g)$.
1.) Is it possible that $f$ is differentiable on any point of the set $M(g)$? I would say no because $M(g)$ doesn't contain inner points.
2.) How do I show that $M(g)$ has no inner points in general?
EDIT
I think I have found a way, this would be a short sketch of it:
Let's assume that $x\in M(g)$ is an inner point. So there must be a neighbourhood $N(x)\subset M(g)$. We choose a $y\neq x$ with $y\in N(x)$ which is path connected to $x$ and $g(x)-g(y) =0$. By the mean value theorem we know that there exists a point $z$ on the path between $x$ and $y$ such that $g(x)-g(y)= Dg(z)(x-y)$. However, this yields $0=Dg(z)(x-y)\implies x=y$ which is a contradiction as we have assumed $x\neq y$ and $Dg$ can't be $\textbf{0}$.
What do you think? Is this correct?
Generally, to restrict a differentiable function $f$ to a differentiable function on a subset, we want that subset to be an immersed or embedded submanifold, i.e. the differential (Jacobian) of the inclusion map is injective. For a level set like the $M(g)$ given to be an embedded submanifold it is sufficient that the differential (Jacobian) of $g$ be surjective everywhere on the set where $g$ vanishes. Since $g$ maps to $\mathbb{R}$, this simply means that there is nowhere in the set $M(g)$ where each partial derivative of $g$ vanishes. Certainly there are many ways in which you can't restrict differentiable functions if $g$ doesn't satisfy this property. Hope this answers part of your question.
Edit after your edit: with your new condition that $Dg \neq 0$ on $M(g)$, $f$ will always be $C^1$ when restricted to $M(g)$, however you are right that $M(g)$ won't contain any balls of $\mathbb{R}^n$, for it will always be of dimension (as a $C^1$-manifold) one less than $n$. However, $f$ will still be differentiable on $M(g)$, it will just be with respect to $M(g)$'s topology and differentiable structure (of dimension $(n-1)$). All of this is shown in any book on smooth manifolds (e.g. Chapter 4 of Lee's Smooth Manifolds book).