Let $U\subset\mathbb{R}^{m+n}$ open set, $f:U\longrightarrow\mathbb{R}^n$, $f\in C^{k}$, and $c\in\mathbb{R}^n$. Set: $$M=\{p\in U; f(p)=c\textrm{ and }f'(p):\mathbb{R}^{m+n}\longrightarrow\mathbb{R}^n\textrm{ surjective}\}.$$ Then $M$ is a open set in $f^{-1}(\{c\})$.
I can see that $M\subset U$ and $M\subset f^{-1}(\{c\})$ so $M\subset U\cap f^{-1}(\{c\})$ where $U\cap f^{-1}(\{c\})$ is a open set in $f^{-1}(\{c\})$, but to show that $M$ is a open set in $f^{-1}(\{c\})$ I don't know if I have to show the equality or have a other way. Somebody have any hint?
In your notation, consider the set $$ V = \{p \in U : \text{$f'(p)$ is surjective (i.e., has rank $n$)}\}. $$ The main point is to show $V$ is open, since $M = f^{-1}(c) \cap V$.
How you accomplish this may depend on how you think of "maximum rank"/surjectivity. One approach is to pick a non-singular $n \times n$ submatrix of $f'(p)$, and to use continuity of the determinant to show there is a neighborhood of $p$ on which this submatrix has non-vanishing determinant.