Let $\mathcal{D}$ denote the Schwarzian derivative. I have to prove that if $\mathcal{D}f(x)$ exists $\forall x$ then $\mathcal{D}f^{-1}$ exists $\forall x\in D_{f^{-1}}$ then find a formula.
I tried changing the variable $x$ into $f^{-1}(x), f(x), f'(x), (f^{-1})'(x)$ and after they all got me nowhere I realised I wasn't doing math anymore but bruteforce. I have the feeling that I'll use $$(f^{-1})'=\frac{1}{f'(f^{-1}(x))}$$ somewhere(which I did in some of my change of variables). Any hints will be appreciated.
For the existence of $\mathcal{D}(f^{-1})$, note that the existence of $\mathcal{D}f$ implies $f'(x) \neq 0$, hence $f$ is a diffeomorphism, $f^{-1}$ is differentiable as often as $f$ is, so sufficiently often for the Schwarzian derivative, and $(f^{-1})'(y) \neq 0$ always. So $\mathcal{D}(f^{-1})$ exists.
By the chain rule for Schwarzian derivatives, we have
$$0 = \mathcal{D}(\operatorname{id}) = \mathcal{D}(f\circ f^{-1}) = \bigl((\mathcal{D}f)\circ f^{-1}\bigr)\cdot \bigl((f^{-1})'\bigr)^2 + \mathcal{D}(f^{-1}).$$
Now it is a simple rearrangement to obtain the a formula for $\mathcal{D}(f^{-1})$.