 Solution

Define the scalar function $H(s)=⟨f(x+s(y-x)),z⟩$ on $[0,1]$. It is continuous there and differentiable inside, with derivative $⟨df(x+s(y-x))(y-x),z⟩$ by the chain rule. The one-dimensional mean value theorem gives $H(1)-H(0)=H^{\prime }(\theta )$ for some $0<\theta <1$, proving the displayed vector-paired version. No claim of a single intermediate point giving a vector equality without pairing is required.

For the final assertion, use the scalar mean value theorem directly on $s\mapsto g(sh)$: it is continuous at $s=0$ and differentiable for $s>0$. Thusfor some $0<{\theta }_h<1$. The assumed limits of all partial derivatives imply $\Vert \nabla g(x)\Vert \to 0$ as $x\to 0$, $x\ne 0$. Consequently $|g(h)-g(0)|/\Vert h\Vert \le \Vert \nabla g({\theta }_{h}h)\Vert \to 0$. Hence $\boxed{dg(0)=0}$ and $g$ is differentiable there. This argument uses only differentiability on the interior of the line segment, so it does not assume the conclusion at its endpoint.