IMC 2002 · Day 1 · P2

Assume that there exists such a function. Since $f'(x) = f(f(x)) > 0$, the function is strictly monotone increasing.

By the monotonity, $f(x) > 0$ implies $f(f(x)) > f(0)$ for all $x$. Thus, $f(0)$ is a lower bound for $f'(x)$, and for all $x < 0$ we have $f(x) < f(0) + x \cdot f(0) = (1 + x) f(0)$. Hence, if $x \le -1$ then $f(x) \le 0$, contradicting the property $f(x) > 0$.

So such function does not exist.