What i have done is very small.
$$f'(x)=f(f(x))\implies f(f'(x))=f(f(f(x)))$$Now $$f(f(f(x)))=f'(f(x))$$Hence$$f(f'(x))=f'(f(x))$$Now i am blank. What to do for the proof
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3I don't know how to solve this problem offhand, but there are several things you've not yet used. One is that the codomain is $0 < f(x) < \infty$, so (i) you can safely divide by $f(x)$ (or cancel $f(x)$ from the two sides of an equality), and (ii) You know that $f'(x) > 0$ for all $x$, so that $f$ must be increasing. You also know $f$ is differentiable, so it's continuous, hence it's monotone continuous, so it must be injective. None of this may be of any use, but it's always risky to ignore the hypotheses. (example: If $0$ was in the codomain, then $f(x) = 0$ would be a solution!) – John Hughes Jul 20 '20 at 13:17
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1In addition to what @JohnHughes mentioned, $f$ is also infinitely differentiable (chain rule) and also convex, since its second derivative $f''(x)=f'(x)f'(f(x))$ is strictly positive. – Vercassivelaunos Jul 20 '20 at 13:23
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1Also: https://math.stackexchange.com/q/2181530, https://math.stackexchange.com/q/1120041, https://math.stackexchange.com/q/848627 – Martin R Jul 20 '20 at 14:01
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Note that under the hypothesis $f$ is increasing. So, $f(f(x))>f(0)$ for all $x\in\Bbb R$. So, $f'(x)$ has a lower bound which is $f(0)$.
Hence $f(x)<f(0)+xf(0)=(1+x)f(0)$ for all $x<0$. So, for $x\leq -1$ we have $f(x)\leq 0$, contradiction.
Sumanta
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@Sumanta in that case $f'(x)> f(0)\implies f(x)>xf(0)+c$. But how $f(x)<f(0)+xf(0)=(1+x)f(0)$ – Sassy Math Jul 20 '20 at 14:05
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