This is the exercise 6 in page 165 of Analysis II of Amann and Escher
Suppose $X$ is open in $\Bbb R^n$, $F$ is a Banach space and $f:X\to F$. Also suppose that $x_0\in X$ and choose $\epsilon>0$ such that $\Bbb B(x_0,\epsilon)\subset X$. Finally let $$x_k(h):=x_0+h_1e_1+\ldots+h_ke_k\quad\text{for }k=1,2,\ldots,n\text{ and }h\in\Bbb B(x_0,\epsilon)$$
Prove that $f$ is differentiable at $x_0$ if and only if for every $h\in\Bbb B(x_0,\epsilon)$ such that $h_k\neq 0$ the limits $$\lim_{\substack{h_k\to 0\\h_k\neq 0}}\frac{f(x_k(h))-f(x_{k-1}(h))}{h_k},\quad k=1,\ldots,n$$ exists in $F$.
First some observations: the vectors $e_k$ are the standard basis of $\Bbb R^n$ and $h:=(h_1,\ldots,h_n)$. It seems a error the definition of $h\in\Bbb B(x_0,\epsilon)$ because its possible that $x_k(h)\notin X$. It seems that it must be $h\in\Bbb B(0,\epsilon)$ instead.
From now, to short things a little bit, I set $x_k:=x_k(h)$. I dont know exactly what to do or what to search for. Searching something for the first implication I search for an useful upper bound to show the existence of the total derivative at $x_0$, by example
$$\frac{\|f(x_0+h)-f(x_0)-\partial f(x_0)h\|}{|h|}=|h|^{-1}\left\|\left(\sum_{k=1}^n(f(x_k)-f(x_{k-1}))\right)-\partial f(x_0)\right\|\\\le\sum_{k=1}^n\frac{\|f(x_k)-f(x_{k-1})-\partial_k f(x_0)h_k\|}{|h_k|}$$
with $|h_k|\neq 0$ for all $k=1,\ldots,n$, and we used the fact that $|h|\ge|h_k|$. But I dont see how to continue from here.
After I thought that if I stablish an equivalence between the existence of the stated limits and the continuity of the partial derivatives of $f$ at $x_0$ Im done. But nothing useful comes from this idea.
Some help will be appreciated.
P.S.: I forget to say that the stated limits are equivalent to state the existence of the partial derivatives $\partial_k f(x_{k-1})$.