 Solution

For $x\ne y$, apply the scalar mean value theorem to $h(t)=⟨f(x+t(y-x)),z⟩$ on $[0,1]$. It is continuous on the closed interval and differentiable inside by the stated segment and regularity assumptions. The chain rule givesThus $h(1)-h(0)=h^{\prime }(\theta )$ for some $\theta \in (0,1)$, precisely the asserted identity. If $x=y$, both sides are zero when the zero-direction derivative is naturally taken as $Df(x)0$.

For the final claim, all partial derivatives tending to zero imply $\Vert \nabla g(x)\Vert \to 0$ away from the origin. For a nonzero small vector $h$, the one-variable function $t\mapsto g(th)$ is continuous on $[0,1]$ by continuity at zero and differentiable on $(0,1)$. Thereforefor some $\theta \in (0,1)$. For any $\epsilon >0$, if $\Vert h\Vert$ is sufficiently small, the entire punctured segment has gradient norm below $\epsilon$. Hence $|g(h)-g(0)|\le \epsilon \Vert h\Vert$, provingThis argument does not presuppose differentiability at the origin.