There is no analytic local solution at $0$ to $f'=e^{f^{-1}}$, $f(0)=0$. The reason is, that such a solution would be extendible to $\mathbb{R}$, which we know to be impossible by Christian Remling previous [answer][1]. For convenience of notation  I shall consider the equivalent equation $g'(x)=e^{g(g(x))}$, satisfied by $g(x):=-f^{-1}(-x)$. 

**1.** The first observation is that the Taylor series of $g$ has non-negative coefficients, which follows immediately by induction equating the  coefficients of $g'$ and $e^{g\circ g}$ (incidentally, this series is the EGF of the positive integer sequence [OEIS A214645][2], as also remarked [here][3]). 

**2.** A general elementary fact: any real analytic function on an interval $I$, whose Taylor series at some point $x_0\in I$ has non-negative coefficients, has Taylor series with non-negative coefficients ay any point $x\in I$, $x\ge x_0$; the reason being: this is clear for $x_1\ge x_0$ within the radius of convergence of $x_0$, and since there is a uniform radius of convergence at any $y\in [x_0,x]$, one reaches $x$ by finitely many steps $x_0<x_1\dots<x_n=x$. 

**3.** Consequence: any analytic local solution of $g'(x)=e^{g(g(x))}$ at $0$ is strictly increasing in the interval where it is defined (even beyond where explicitly said by the equation, that is $\operatorname{dom}(g\circ g)$ !). We can therefore perform the following  construction: 

**4.** Let $g_0$ be an analytic local solution and assume $[0,a)\subset\operatorname{dom}(g_0)$ for some $a_0>0$ :
$$\begin{cases} g_0(0)=0,  \\ g_0'(x)=e^{g_0(g_0(x))}, & \quad 0\le x< g_0^{-1}(a_0)\, . 
 \end{cases}$$
Then, putting $a_1:=g_0(a_0)$,  $$g_1(x):=\log g_0'(g_0^{-1}(x))$$  
defines a local solution that extends $g_0$ to the interval $[0,a_1)$. 

**5.** Conclusion: if there is an analytic local solution $g_0$, we can apply iteratively the preceding construction and  define a sequence $g_n:[0,a_n)\to\mathbb{R}$ of successive extensions with $a_n:=g_{n-1}(a_{n-1})$. Putting $a:=\sup_n a_n$, and gluing the $\{g_n\}_n$  together one has a solution $g:[0,a)\to\mathbb{R}$. Of course, since  $g(0)=0$ and $g'(x)>g'(0)$ for any $x>0$,  $g$ has no fixed points and $a=+\infty$, a contradiction as there are no solutions on $[0,\infty)$. 

[1]:http://mathoverflow.net/questions/258525/how-do-i-solve-this-displaystyle-f-ef-1/258544#258544
[2]:http://oeis.org/A214645
[3]:http://mathoverflow.net/questions/258525/how-do-i-solve-this-displaystyle-f-ef-1/258695#258695