In this MIT open course video Denis Auroux said that given a function $f(x, y)$ and a restriction level surface $g(x, y) = c$, for any vector $u$ tangent to this surface $g=c$, we must have:
$$ \left.\frac{df}{ds} \right|_{u} = 0, $$
where $s$ is a parameter such that
$$ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x_0 \\ y_0 \end{pmatrix} + s \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} $$
making the position vector $ \begin{pmatrix} x \\ y \end{pmatrix} $ moving the direction of $u$.
I am a little suspicious of this conclusion. Suppose $g$ is just a plane $x - y = 0$ and $f(x, y) = x^2 + y^2$, I can imagine a lot of vectors on the surface can be found as counter examples. But I am not so confident of my own reasoning so far. Can anyone explain?
He's not saying that this holds everywhere on the surface: simply that it holds at the optima. If the derivative is positive, we could find a small positive value of $s$ such that the corresponding point has $f(Q) > f(P)$ and a small negative value of $s$ such that $f(Q) < f(P)$, therefore $P$ cannot be a local optimum of $f$ restricted to the curve traced out by varying $s$ (and consequently, can't be a local optimum of $f$ restricted to the surface). Similarly if the derivative were negative.
However, if $df/ds = 0$ at $P$, then $P$ can be a local optimum. (but it still might not be)
Compare with the one-dimensional version: the optima of a differentiable function on $(a,b)$ can only occur at the points where $f'(c) = 0$.