If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative

If $m=0$, then $a=0$, $U=\{0\}$, and the only increment is $h=0$. The unique linear map $L:{\mathbb{R}}^0\to {\mathbb{R}}^n$ has $L0=0$, so $f(a+h)-f(a)-Lh=0$ for every admissible increment. It satisfies the normwise remainder estimate with any $\delta >0$. There are no coordinate directions; $Jf(a)$ is the empty-column $n\times 0$ matrix of $L$. Thus the conclusion holds without dividing by $m$ or invoking a punctured limit at an isolated point.

If $m\ge 1$ and $n=0$, every value of $f$ is the unique zero vector. The unique linear map $L:{\mathbb{R}}^m\to {\mathbb{R}}^0$ therefore leaves zero remainder for every increment, and its matrix, like $Jf(a)$, has zero rows. This proves the conclusion directly, without applying the positive-codomain vector mean-value inequality. For the rest of the proof assume $m,n\ge 1$.

Choose a ball around $a$ on which the partial derivatives exist. Given $\epsilon >0$, continuity at $a$ gives a smaller ball on which every $\mathrm{\parallel }{\mathrm{\partial }}_{j}f(z)-{\mathrm{\partial }}_{j}f(a){\mathrm{\parallel }}_2<\epsilon \mathrm{/}m$.

For $h$ in that smaller ball, [L1] writes the increment as coordinate segments. On each segment apply [L2] to the one-variable map obtained after subtracting the fixed linear term $Jf(a)$; its derivative norm is at most $\epsilon \mathrm{/}m$.

Summing the segment bounds gives $\mathrm{\parallel }f(a+h)-f(a)-Jf(a)h{\mathrm{\parallel }}_2\le (\epsilon \mathrm{/}m){\sum }_j\mathrm{\mid }{h}_j\mathrm{\mid }\le \epsilon \mathrm{\parallel }h{\mathrm{\parallel }}_2\mathrm{/}\sqrt{m}\le \epsilon \mathrm{\parallel }h{\mathrm{\parallel }}_2$. Since $\epsilon$ is arbitrary, this verifies the normwise remainder condition and $Jf(a)$ is $Df(a)$. Together with the two zero-dimensional cases this proves the full Statement.