What is solution of $f'(f(x))=\exp(f'^{-1}(x))$ with $f'^{-1}$ is a compositional inverse of $f'$?

,Assume $$f$$ a bijective and differentiable function on its domain , I want to find solutions of this functional such that $$f\colon\mathbb{R} \to \mathbb{R}$$; $$f'(f(x))=\exp(f'^{-1}(x))$$ such that $$f'^{-1}$$ is a compositional inverse of $$f'$$ , I have tried $$f(x)=-x$$ seems works because we have $$f'(f(x))$$ is increasing function in the same time $$f^{-1}$$ exist and would be increasing imply $$f'^{-1}$$ exist and increasing , if we raise exponential we have increasing functions, but am not sure about this solution ? Is there a clear solution, because its seems that it has a nontrivial solution which it is a formal power series arround $$x=0$$ , we may get its coefficients using inverse function theorem ?

• Should $f'^{-1}$ be the inverse of $f$ or of $f'$? You say two different things. Your attempt seems to suggest $f'^{-1}$ is the inverse of $f$, but the notation suggests otherwise. – Batominovski Apr 26 at 19:49
• Compositional inverse of f' , just a wrong typo – zeraoulia rafik Apr 26 at 19:50