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

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 $I times I backslash cup_{x in I} {(x,x)}$