Cartan Differentiable calculus. Show $g(x,y)= \frac{f(x)-f(y)}{x-y}$ is differentiable at $(x_{0},x_{0})$

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I'm doing problem 8 of Cartan Differentiable Calculus book. The problem says as follow:

Let $f$ assume its values in a Banach space $E$, an let it be of class $\mathcal{C}^1$ in an open interval $I$. Put

$ \begin{cases} g(x,y)&= \frac{f(x)-f(y)}{x-y} ~ \text{ if } x \neq y\\ g(x,x)&= f'(x) \end{cases} $

If $f''(x_{0})$ exists at $x_{0} \in I$ show that $g$ is differentiable in $(x_{0},x_{0})$

So, I think that we should have

$Dg(x_{0},x_{0})[h_{1},h_{2}] = \frac{f''(x_{0})}{2}(h_{1}-h_{2})$

but I've manage nothing more. The version I have says the following hint:

Apply the mean value theorem to the function

$ f(x)=xf'(x_{0})- \frac{(x-x_{0})^2}{2}f''(x_{0}) $

I think there is a typo and it should be $h(x)=...$

Either way I would gladly appreciate any help.

Edit:

I already proved that g is continous in $I \times I$ and its $\mathcal{C}^1$ in $I \times I \backslash \cup_{x \in I} \{(x,x)\}$