Suppose that a function $f:\Omega \subset \mathbb R^N \to \mathbb R$ has a maximum point $x_0 \in \partial \Omega$ (where we have assumed $\Omega$ smooth and bounded$).
Is it true that the Neumann derivative is zero?
$$\frac{\partial f}{\partial \nu}(x_0) = 0$$
Clearly if $x_0$ is in the interior of $\Omega$, the first derivatives are zero.
Take $\Omega = B_1(0)$, the unit ball of $\mathbb{R}^n$ and let $f(x) = ||x||^2 = x_1^2 +...+x_n^2$. Then, $f$ has maximum at all points of $\partial \Omega$, and for the gradient we have $\nabla f(x) = 2x$. Since at $x_0\in \partial \Omega$ the unit (inward) normal is $-x_0$, then $\partial_\nu f(x_0) = -2 x_0 \cdot x_0 = -2$.