Problem: Prove that there is no differentiable function from $\mathbb{R} \to (0, \infty)$ satisfying $f'(x)=f(f(x))$.
I could not make much progress, except for observing that any derivatives (any order, whenever they exist) are always positive.
Please help. Even hints are appreciated.
Since $f$ outputs positive numbers, $f'$ must be positive. Since $f'=f\circ f$, then $f''$ exists, and $f''=\left(f'\circ f\right)\cdot f'$, and you can see that $f''$ is also positive. With $f$, $f'$, and $f''$ positive on $\mathbb{R}$, $f$ must have a horizontal asymptote $y=c\geq0$ as $x\to-\infty$ and $\lim_{x\to-\infty}f'(x)=0$. (Formal proof below.) Then $$f(c)=\lim_{x\to-\infty}f(f(x))=\lim_{x\to-\infty}f'(x)=0$$ and it is not permitted that $f(c)=0$.
With $f$ and $f'$ positive, $f$ is bounded below and increasing. This guarantees $c$ exist, and gives the existence of the horizontal asymptote. But it isn't automatic that $\lim_{x\to-\infty}f'(x)$ exists. For example, maybe the curve is flat as it moves left, and occasionally, for very brief periods, takes on a very steep slope. This is why (as Greg Martin notes in the comments) we must make use of $f''$ being positive.
The logic is the same with $f'$ as was with $f$. Since $f'$ and $f''$ are positive, then $f'$ has a left asymptote at $y=d\geq0$. But $d$ must equal $0$ since otherwise $f$ wouldn't be bounded below.