Let $f:\mathbb{R}^n\to \mathbb{R}$ be a differentiable function, with $a\in \mathbb{R}$ a regular value of $f$.
Let $M=f^{-1}((-\infty,a])$. Then $M$ is an $n$-manifold with boundary, whose boundary is $\partial M=f^{-1}(a)$.
Let $p\in \partial M$. Then $T_p\partial M=\{\nabla f(p)\}^{\perp}$.
Question: Does $\nabla f(p)$ point outwards of $\partial M$?
I know it's not true if we take the interval the other way around. E.g. $-1$ is a regular value of $f:\mathbb{R}^2\to \mathbb{R}$, $f(x,y)=-x^2-y^2$, but the gradient of $f$ points inwards of the boundary of $f^{-1}([-1,+\infty))$ which is $S^1$.
In fact, by symmetry, if the answer to the question is positive, I suspect that if we take $M=f^{-1}([a,+\infty))$ then $\nabla f(p)$ always points inwards of $\partial M$.
Best Answer
By assumption $f(p)=a$ and $\nabla f(p)=: n\ne0$. For $X\in T_p$ we have $$f(p+X)-f(p)=n\cdot X+o\bigl(|X|\bigr)\qquad(X\to0)\ .$$ Now put $X:= \lambda n$ with $\lambda>0$. Then $|X|=\lambda|n|$ and therefore $$f(p+\lambda n)-f(p)=\lambda|n|^2+o(\lambda|n|)=\lambda |n|^2\bigl(1+o(1)\bigr)\qquad(\lambda\to0+)\ .$$ It follows that $f(p+\lambda n)>f(p)=a$ for all suitably small $\lambda>0$, which implies that $n$ points to the outside of $M$.