= Solution
For $x\ne y$, apply the scalar <mean value theorem> to $h(t)=\langle f(x+t(y-x)),z\rangle$ on $[0,1]$. It is <continuous> on the closed interval and differentiable inside by the stated segment and regularity assumptions. The <chain rule> gives
$$
h'(t)=\langle Df(x+t(y-x))(y-x),z\rangle.
$$
Thus $h(1)-h(0)=h'(\theta)$ 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 $\|\nabla g(x)\|\to0$ 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 $(0,1)$. Therefore
$$
g(h)-g(0)=\nabla g(\theta h)\cdot h
$$
for some $\theta\in(0,1)$. For any $\varepsilon>0$, if $\|h\|$ is sufficiently small, the entire punctured segment has <gradient> <norm> below $\varepsilon$. Hence $|g(h)-g(0)|\leq\varepsilon\|h\|$, proving
$$
\boxed{Dg(0)=0}.
$$
This argument does not presuppose <differentiability> at the origin.
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