This question is a spin-off of this one, in which the OP asks whether there is a solution $f:\mathbb R\to\mathbb R$ of the functional equation (not exactly an ODE) $f'=e^{f^{-1}}$, where $f^{-1}$ is the compositional inverse of $f$. The posted answer exploits the growth of $f(x)$ when $x\to-\infty$ and obtains a contradiction, which resolves the question nicely, but also invites the following question: what if we restrict to $f:\mathbb R_{\ge0}\to\mathbb R_{\ge0}$ and impose $f(0)=0$? This idea has been explored in the comments, where a formal power series expansion is obtained for $f$ which does not seem to converge for any $x\ne0$.

Taking another approach, we can use an iteration scheme starting from $f_1(x)=x$ and inductively solve the ODE $f_{n+1}'=e^{f_n^{-1}}$ with the initial condition $f_{n+1}(0)=0$ to obtain $f_{n+1}$, much in the spirit of Picard iteration. Explicitly, for example, we have

$f_2'=e^x$ and $f_2=e^x-1$;

$f_3'=e^{\ln(x+1)}=1+x$ and $f_3=x+x^2/2$;

$f_4'=e^{\sqrt{1+2x}-1}$ and $f_4=e^{\sqrt{1+2x}-1}(\sqrt{1+2x}-1)$

and the next iteration produces non-elementary functions. It is clear that the sequence $(f_{2k-1})_{k\ge1}$ is increasing, $(f_{2k})_{k\ge1}$ is decreasing, and $f_{2k-1}<f_{2k}$, so there are respective limits $f_-=\lim_{k\to\infty} f_{2k-1}$ and $f_+=\lim_{k\to\infty} f_{2k+1}$, with $f_-\le f_+$. It is also clear that from $n\ge2$ on the function $f_n'=e^{f_{n-1}^{-1}}$ is positive and increasing, so $f_n$ is increasing and convex, which can be passed to the limit to show that both $f_-$ and $f_+$ are also increasing and convex. As such they are continuous, and by Dini's theorem $f_{2k-1}$ converges to $f_-$ locally uniformly and similarly for $f_+$. Furthermore, the inequality $|x-y|\le |f_n(x)-f_n(y)|$ (as $f_n'=e^{f_{n-1}^{-1}}\ge1$) can also be passed to the limit. Then the following chain of inequalities:

$|f_-^{-1}(x)-f_{2k-1}^{-1}(x)|\le |x-f_-(f_{2k-1}^{-1}(x))|=|f_{2k-1}(f_{2k-1}^{-1}(x))-f_-(f_{2k-1}^{-1}(x))|$

shows that $f_{2k-1}^{-1}$ converges locally uniformly to $f_-^{-1}$, which then implies $f_{2k}'$ converges locally uniformly to $e^{f_-^{-1}}$. Hence $f_+'=e^{f_-^{-1}}$, and similarly $f_-'=e^{f_-^{-1}}$. From this it can be shown that $f_{2k-1}$ converges to $f_-$ locally in $C^\infty$, so both $f_-$ and $f_+$ are smooth functions, and they form an orbit of order at most 2 of the above iteration scheme. Moreover it can be shown that the first $n$ terms of the Taylor expansion of $f_n$ agrees with what have been calculated formally in the previous comments, so both $f_-$ and $f_+$ have the same Taylor expansion as calculated using formal power series expansion.

In light of the above, *a priori* the following three scenarios can happen:

- $f_-\neq f_+$ and we have a genuine orbit of order 2, consisting of two functions having the same Taylor expansion at 0 but not being identical.
- $f_-=f_+$ is an actual solution to the equation $f'=e^{f^{-1}}$, but it is merely $C^\infty$ but not analytic, having a divergent power series expansion at 0.
- $f_-=f_+$ is an actual solution to the equation $f'=e^{f^{-1}}$, and it is analytic on a neighborhood of 0; we are just misled by the first 100 or so terms of the Taylor expansion.

Now finally comes the question: which of the above scenario is the reality? In the first two scenarios, one can also ask what is the growth rate of $f_-(x)$ and $f_+(x)$ as $x\to+\infty$.