How do you solve $f'(x) = f(f(x))$?

A friend told me to solve the following differential equation:

$$f'(x)=f(f(x))$$

I have no idea how to solve this! This doesn't seem to be an ordinary differential equation and I can't even solve this numerically!

I think my friend is trolling me.

This question has been asked on MathOverflow, this answer is migrated from there: -

There are two closed form solutions:

$$\displaystyle f_1(x) = e^{\frac{\pi}{3} (-1)^{1/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$ $$\displaystyle f_2(x) = e^{\frac{\pi}{3} (-1)^{11/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$

The solution technique can be found in this paper.

For a general case, solution of the equation

$$f'(z)=f^{[m]}(z)$$

has the form

$$f(z)=\beta z^\gamma$$

where $\beta$ and $\gamma$ should be obtained from the system

$$\gamma^m=\gamma-1$$ $$\beta^{\gamma^{m-1}+...+\gamma}=\gamma$$

In your case $m=2$.

Note: I merely brought this to your attention, all credit should go to Anixx on MO.