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I'm interesting to know if the below equation has a closed form solution , but it seems to me that has unic solution which it's power series , then my question here is :

Question: let $f$ be a function defined as :$f:\mathbb{R}\to \mathbb{R}$

Could be this : $\displaystyle f'=e^{f+f^{-1}}$ has closed form and does it has a local solution at $0$?

Note 01: $f^{-1}$ is the compositional inverse of $ f$

Note 02: The motivation of this question is to check the rate growth of the titled equation

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    Do you want to impose the additional condition $f(0)=0$? – 2017-02-16
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    I don't believe there is a closed form. Now to the point: if (say) $f(x)$ grows faster than $x$, then $f^{-1}(x)$ grows slower, so pretty soon it ceases to be relevant in comparison. Then $f'=e^f$ (which, BTW, **does** have a closed-form solution) becomes a pretty adequate approximation for studying the growth rate. – 2017-02-16
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    and f'=e^f is signed http://oeis.org/A214645 – 2017-02-16
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    May be you find this [link](http://mathoverflow.net/questions/258611/f-ef-1-again?newsletter=1&nlcode=688348%7cf0c6) useful. – 2017-02-16
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    it is my referenced question posted again by zheng , but this question is different and i ask about the sum of f^-1+ f in exponent of exp function – 2017-02-16

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