I'm trying to prove the following:
Show that $f$ grows fastest along path for which $\gamma'(t)=\nabla f(\gamma(t))$ than along any other path.
My reasoning is that of course $f$ along $\gamma$ must be growing the fastest since by definition the gradient at a point $x_0$ is the direction to which the function grows fastest and its magnitude gives the steepness of the growth.
But what about "any other path"?
Let $\eta$ be any other path. Then the growth of $f$ along $\eta$ is measured by the the derivative: $(f\circ \eta)'(t)=\nabla f(\eta(t))\cdot\eta'(t)$, which by the Cauchy-Schwarz inequality is $\leq || \nabla f(\eta(t)||\ || \eta'(t)||$. Therefore, the maximum value of the derivative will be attained when this is an equality, i.e. when the vectors are parallel. In other words, when $\eta'(t)=\lambda \nabla f(\eta(t))$, for some $\lambda$.
However, note that you can always reparametrize $\eta$ to get $\lambda=1$.