Let $f$ be a function such that :$f:\mathbb{R}\to \mathbb{R}$ and $f^{-1}$ is a compositional inverse of $f$. I would'd like to know how do I solve this class of differential equation : $$\displaystyle \ f'= e^{\displaystyle {f}^{-1}}?$$ 


**Note 01:** $f' =\displaystyle\frac{df}{dx}$.


**Edit:** ${f}^{-1}$ is the inverse compositional of $f$, for example $\log$ is the inverse application of exp function .


**Note 02**: I have edited my question to clarify the titled question that related to ${f}^{-1}$


Thank you for any help 



How do i solve this : $\displaystyle \ f'=e^{{f}^{-1}}$?
There is no such function. Since $f$ would have to map $\mathbb R$ *onto* $\mathbb R$ for the equation to make sense at all $x\in\mathbb R$, it follows that $f^{-1}(x)\to -\infty$ also as $x\to -\infty$, so $f'\to 0$. Thus $f(x)\ge x$, say, for all small enough $x$, hence $f^{-1}(x)\le x$ eventually, but then the equation shows that $f'\le e^x$, which is integrable on $(-\infty, 0)$, so $f$ would approach a limit as $x\to -\infty$ and not be surjective after all.


