Let $A \subset \mathbb{R}$ be open, $f,g: A \rightarrow \mathbb{R}$ class $C^1$. Let $x_0 \in A$ satisfy $g(x_0) = 0$ and define $S = g^{-1}\{0\}$, also $\nabla g(x_0) \neq 0.$ If $f\lvert_{S}$ has a maximum or minimum at the point $x_0$ there exists $\lambda \in \mathbb{R}$ such that $$\nabla f(x_0) = \lambda \nabla g(x_0)$$
Proof:
For every path $c$ through $x_0$ such that $x_0 = c(0)$ it follows:
$$g(c(t)) = 0 \implies \frac{d}{dt}g(c(t)) = 0 \implies Dg(c(t))c'(t)=0$$
therefore $$(\nabla g(x_0)\vert c'(0)) = 0$$
So $\nabla g(x_0)$ is orthogonal to every tangent vector of the curve on $g(x)=0$ at the point $x_0$.
And the proof goes on, and it's all good from there, but I'm just not sure what I'm missing in the first implication, why does it follow that $\frac{d}{dt}g(c(t)) = 0$
I have a feeling it's something simple, but I just can't see it right now. Thanks in advance!
It's right there in the question. $g(c(t)) = 0$ for all $t$, so in particular it is a constant function with derivative $0$.