How do i solve this : $\displaystyle \ f'=e^{{f}^{-1}}$? 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 
 A: A formal Taylor series (e.g.f.) solution about the origin can be obtained a few ways.
Let $f^{(-1)}(x) = e^{b.x}$ with $(b.)^n=b_n \;$ and $ \; b_0=0$.
Then A036040 (Bell polynomials) gives the e.g.f. 
$$e^{f^{(-1)}(x)}= e^{e^{b.x}}= 1 + b_1 x + (b_2+b_1^2) \frac{x^2}{2!}+(b_3+3b_1b_2+b_1^3)\frac{x^3}{3!}+\cdots \; ,$$
and the Lagrange inversion / series reversion formula (LIF) A134685 gives
$$f'(x)= \frac{1}{b_1} + \frac{1}{b_1^3} (-b_2) x + \frac{1}{b_1^5}(3b_2^2-b_1b_3)\frac{x^2}{2!}+\cdots \; .$$
Equating the two series and solving recursively gives
$$b_n \rightarrow (0,1,-1,3,-16,126,-1333,...)$$
which is signed A214645. This follows from the application of the inverse function theorem (essentially the LIF again)
$$f'(z) = 1/f^{(-1)}{'}(\omega) \; ,$$
when $(z,\omega)=(f^{(-1)}(\omega),f(z)) $, leading to
$$f^{(-1)}{'}(x) = \exp[-f^{(-1)}(f^{(-1)}(x))],$$
the differential equation defining signed A214645.
Applying the LIF to the sequence for $b_n$ gives the e.g.f. $f(x)=e^{a.x}$ equivalent of F.C.'s o.g.f.
$$ a_n \rightarrow (0,1,1,0,1,-6,52,...).$$
As another consistency check, apply the formalism of A133314 for finding the multiplicative inverse of an e.g.f. to find the e.g.f. for $\exp[-A(-x)]=\exp[f^{(-1)}(x)]$ from that for 
$$\exp[A(-x)]= 1 - x + 2 \frac{x^2}{2!}-7 \frac{x^3}{3!}+\cdots \; ,$$
which is signed A233335, as noted in A214645. This gives $f'(x)=a. \; e^{a.x}$.
A: 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.
A: I have an idea to express solutions of transcend equations. If you replace functions in algebraic equations by operators then this way will suit operator equations.
You can find my post in mathoverflow by User 'qian' or 'tag' symbolic computation. 
In my post I give  the equation $x+e^{x}=0$ an explicit solution:
$x=\Big(I_{1}\{C^{3}_{1,3}[A^{3}_{1,2}(\varphi_{a})]C^{3}_{2}[A^{3}_{2,3}(\varphi_{p})]\}\Big)(0,e)$:
Your equation is $f'=e^{f^{-1}}$, we denote $f'=\alpha(f)$,$f^{-1}=I(f)$,that is say we consider differential and inverse as special unary operators.
 We denote subtraction and power as $\varphi_{s}$,$\varphi_{p}$ respectively as I used in my post then
$f'=e^{f^{-1}}\Longleftrightarrow \varphi_{s}\{\alpha(f),\varphi_{p}[e,I(f)]\}=0$,
The solution will be:
$f=\Bigg(I_{1}\Big\{C^{3}_{1,3}\Big[\Big(A^{3}_{1,2}\{\varphi_{a}C^{2}_{1}[A^{2}_{1}(\alpha)]\}\Big)C^{3}_{2}\Big(A^{3}_{2,3}\{\varphi_{p}C^{2}_{2}[A^{2}_{2}(I)]\}\Big)\Big]\Big\}\Bigg)(0,e)$
This result is only a reference but I hope it will help you.
