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.

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 $f'(x)=a. \; e^{a.x}$.