Suppose I have functions $f,g$ which are strictly concave and strictly increasing. Also the first and second derivatives of $f$ and $g$ exist. This means that $f'(x)>0$ and $g'(x)>0$, and $f''(x)<0$ and $g''(x)<0$. I want to construct a function $h$ such that $g(x)=h(f(x))$. Now I have to prove that $h''(x)$ exists and $h$ is an increasing function.
Furthermore I know that $$-\frac{g''(x)}{g'(x)}\geq -\frac{f''(x)}{f'(x)}$$
I took $h(x)=g(f^{-1}(x))$. This is increasing since $h'(x)=g'(f^{-1}(x)) \cdot (f^{-1})'(x)$. Note that $f^{-1}(x)$ exists and because $f'(x)>0 \Rightarrow (f^{-1})'(y)>0$ and $g'(z) >0$ for all $z \in \mathbb{R}^+$.
I want to apply a similar method to show that the second derivative of $h$ exists, but I do not know how to show $(f^{-1})''(x)$ exists. I know that since the functions $f$ and $g$ are strictly concave we have that $f''(x)<0$ and $g''(x)<0$ but I do not see how this helps here.
Could you please help me?
Note: edit was to update question, I noticed I made a mistake in the formulation.