Let $A \subset \mathbb{R}^n$, with $A$ open, and let $\emptyset\ne S \subset A$. Let $f\in C^1 (A, \mathbb{R} )$ be a function. Let $\vec{a}$ be a point in $S$, and suppose that $\vec{a}$ is a local extremum point for $f \mid S$. Prove that the gradient vector $( \nabla f ) ( \vec{a} )$ is orthogonal to every vector $\vec{v}$ which is tangent to $S$ at $\vec{a}$.
Aside (definition of a vector tangent to a set):
Let $\emptyset\ne S\subset\mathbb{R}^n$ and let $\vec{a}\in S$. A vector $\vec{v} \in \mathbb{R}^n$ is said to be tangent to $S$ at $\vec{a}$ when there exists a differentiable path $\gamma : I \to \mathbb{R}^n$, where $I \subseteq \mathbb{R}$ is an open interval containing $0$, such that the following conditions are fulfilled:
(i) $\gamma (t) \in S$ for every $t \in I$;
(ii) $\gamma (0) = \vec{a}$;
(iii) $\gamma ' (0) = \vec{v}$.
The approach I used is as follows:
Let $u:I\to\mathbb{R}$ be a function, such that $u(t) = f(\gamma(t)), t\in I$. Then $u$ is differentiable, with
$$u'(t)=\langle (\nabla f(\gamma(t)),\gamma'(t) \rangle$$ $$D_{\vec{v}}(\vec{a})=u'(0)=\langle (\nabla f(\gamma(0)),\gamma'(0) \rangle=\langle \nabla f(\vec{a}), \vec{v} \rangle$$
Now I'm wondering why $D_{\vec{v}}(f(\vec{a}))$ should be equal to $0$. This must somehow be related to the fact that $\vec{a}$ is a local extremum of $f$. But how exactly does this make $D_{\vec{v}}(\vec{a})=0$?
Because $a$ is a extremum point for the function $f$, hence for all curves $\gamma:(-\epsilon,+\epsilon) \to S $ such that, $\gamma(0)=a$ and $\gamma^{'}(0)=v$; $0$ is an extremum point for the function $f(\gamma(t))$. Now, because $0$ is an extremum point for this function; by the first derivative test $$\frac{d}{dt}\bigg\rvert_{t=0}f(\gamma(t)) = 0 $$ But as you were trying to argue(correctly), the LHS is just $D_{v}(a)$. Hence, you get that the directional derivative is zero everywhere.
PS: The result is intuitively obvious once you try to remember that the directional derivative is essentially just the $classical$ derivative along a curve passing through your point. Now, if a point is maximum or minimum for a function defined on your surface, it has to be that for any such curve as well, and therefore the $classical$ derivative should be zero at the point on the curve as well. I hope, I have clearly explained it. Feel free to comment if you want further clarification.