A fierce differential-delay equation: df/dx = f(f(x)) Consider the following set of equations:
$$
\begin{array}{l}
y = f(x) \\
\frac{dy}{dx} = f(y)
\end{array}$$
These can be written as finding some differentiable function $f(x)$ such that 
$$
f^{\prime} = f(f(x))
$$ 
For example, say $y(0) = 1$. Then $\left. \frac{dy}{dx} \right|_{x=0}$ is determined by the value of $y(1)$.  The derivative at the $x=0$ had better be negative, otherwise by the time the function gets to 1, the value will be too great and will contradict the alleged value of hte derivative at $x=0$.
Many years ago I tried various techniques to find a solution (other than the trivial $f(x) = 0$) to this equation. It has properties akin to a delay equation, but the delay is variable and strongly depends on the solution itself. I tried expanding as a series; that fails spectacularly. I tried eigenvalue tricks, without any notable progress. Fourier analysis looks good, until you contemplate things like $\sin( \sin( \ldots \sin(x)) \ldots)$ that emerge and that made me give up.  I still think if there are solutions there will be periodic solutions.l
I suspect solving this problem is hard, but perhaps somebody can prove that no non-trivial solution exists. 
Edit after seeing the good complex-valued solution provided by JJaquelin:  
Can anybody find a *real differentiable $f(x)$ other than the trivial $f(x)=0$ that satisfies the conditions, or prove that no such function exists.
 A: This is not the a complete answer. Only two solutions are shown below. Furthermore, I plugged them into the initial equation and it's agree.
$$f'(x)=f(f(x))$$
Search of solutions on the form $f(x)=a x^b$
$$a b x^{b-1} = a(a x^b)^b = a^{b+1} x^{b^2} \rightarrow \begin{cases}b^2 = b-1\\
a^{b+1} = a b \rightarrow a = b^{1/b}\end{cases}$$
$$b=\frac{1}{2}\pm i\frac{\sqrt{3}}{2}; \frac{1}{b}=\frac{1}{2}\mp i\frac{\sqrt{3}}{2}; a=\left(\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\right)^{\frac{1}{2}\mp i\frac{\sqrt{3}}{2}}$$
First solution: $f(x)=\left(\frac{1}{2}+ i\frac{\sqrt{3}}{2}\right)^{\frac{1}{2}- i\frac{\sqrt{3}}{2}}x^{\frac{1}{2}+ i\frac{\sqrt{3}}{2}}$
Second solution: $f(x)=\left(\frac{1}{2}- i\frac{\sqrt{3}}{2}\right)^{\frac{1}{2}+ i\frac{\sqrt{3}}{2}}x^{\frac{1}{2}- i\frac{\sqrt{3}}{2}}$
A: It looks like the contraction mapping theorem can be applied here to guarantee a local solution. Consider 
$$Af(x)=b+\int_0^x f(f(t))dt,\qquad x\in[-a,a]$$
defined on $C^1[-a,a]$, endowed with norm $\|f\|=\|f\|_\infty+\|f'\|_\infty$ (or  any other equivalent norm). Then it's not hard to show that if $B$ is the ball $B=\{f\in C^1[-a,a]: \|f\|\le a\}$, then for small enough $a$ and small enough $b$ (depending on $a$) $A:B\to B$, and the operator $A^2$ is a contraction on $B$. This guarantees the existence and the uniqueness of a solution in $B$, satisfying the initial condition $f(0)=b$.
A: Just playing around a bit
with JJacquelin's answer.
Regarding
$f(x)
=\left(\frac{1}{2}+ i\frac{\sqrt{3}}{2}\right)^{\frac{1}{2}- i\frac{\sqrt{3}}{2}}x^{\frac{1}{2}+ i\frac{\sqrt{3}}{2}}
$,
since
$\frac{1}{2}+ i\frac{\sqrt{3}}{2}
=e^{i\pi/3}
$
and
$\frac{1}{2}- i\frac{\sqrt{3}}{2}
=e^{-i\pi/3}
$,
this becomes
$\begin{array}\\
f(x)
&=\left(e^{i\pi/3}\right)^{e^{-i\pi/3}}x^{e^{i\pi/3}}\\
&=e^{i(\pi/3)e^{-i\pi/3}}x^{e^{i\pi/3}}\\
&=e^{i(\pi/3)(\cos(\pi/3)-i\sin(\pi/3)}x^{\cos(\pi/3)+i\sin(\pi/3)}\\
&=e^{(\pi/3)(i\cos(\pi/3)+\sin(\pi/3)}x^{\cos(\pi/3)+i\sin(\pi/3)}\\
&=e^{(\pi/3)\sin(\pi/3)}e^{(\pi/3)(i\cos(\pi/3)}x^{\cos(\pi/3)+i\sin(\pi/3)}\\
&=e^{(\pi/3)\sin(\pi/3)}e^{(\pi/3)(i\cos(\pi/3)}x^{\cos(\pi/3)}x^{i\sin(\pi/3)}\\
&=e^{(\pi/3)\sin(\pi/3)}x^{\cos(\pi/3)}e^{(\pi/3)(i\cos(\pi/3)}x^{i\sin(\pi/3)}\\
&=e^{(\pi/3)\sin(\pi/3)}x^{\cos(\pi/3)}e^{(\pi/3)(i\cos(\pi/3)}e^{i\ln(x)\sin(\pi/3)}\\
&=e^{(\pi/3)\sin(\pi/3)}x^{\cos(\pi/3)}e^{i(\ln(x)\sin(\pi/3)+\cos(\pi/3))}\\
&=e^{(\pi/3)\sin(\pi/3)}x^{\cos(\pi/3)}(\cos(\ln(x)\sin(\pi/3)+\cos(\pi/3))+i\sin(\ln(x)\sin(\pi/3)+\cos(\pi/3)))\\
\end{array}
$
Don't know if 
this is any use,
but, hey.
