I am reading this chapter about manifolds here and the author writes (page 2):
There is a very important restriction we impose on this situation. It is motivated by our recognition from p. 2–43 that $\nabla g(x)$ is a vector which should be orthogonal to $M$ at the point $x \in M$. If $M$ is truly $(n−1)$-dimensional, then this vector $\nabla g(x)$ should probably be nonzero.
Here $M$ is a level set of $g$.
Why does $M$ is $n-1$-dimensional imply that the gradient of $g$ should be nonzero?
Suppose $M$ is not entirely $(n-1)$-dimensional. That is, suppose there's at least one point $x_0\in M$ such that there's an open $n$-ball $U\subset M$ containing $x_0$. Then the single "restriction" imposed by $g$ in the definition of $M$: $$M=\{x\in \mathbb{R}^n: g(x)=c\}$$ has failed to actually remove one "degree of freedom" at $x_0$. In other words, there can only be an $n$-ball in $M$ surrounding $x_0$ if going in any direction whatsoever doesn't change the value of $g$, i.e., every partial derivative of $g$ is $0$, i.e. $\nabla g(x_0)=0$.
Thus, if we want to ensure there are no such points $x_0$ for any choice of $c$, then we should require that $\nabla g(x)\neq 0$ for all $x\in\mathbb{R}^n$. (And it is not hard to reverse the argument to show this is sufficient as well.)