Functional Equation $f(f'(x))=-f(x)$ Assume $f:\mathbb{R}_{>0}\to\mathbb{R}$ is differentiable and satisfies $\forall x>0:f(f'(x))=-f(x)$. What is $f(x)$?
I know that $f(x)=\ln x$ is a solution, but I don't know if there is another one, or how can I prove that.

It seems the following. 
Differentiate both parts of the equation: 
$$f’(f’(x))f’’(x)=-f’(x).$$ 
Then $$f’(f’(x))=-\frac{f’(x)}{ f’’(x)}.$$
Take $f$ of both parts: 
$$f(x)=-f(f’(x))=f(f’(f’(x)))=f\left(-\frac{f’(x)}{ f’’(x)}\right).$$
Since the function $f$ is injective,  
$$x=-\frac{f’(x)}{ f’’(x)}.$$
This is a differential equation. Solve it 
$$xf''(x)+f'(x)=0$$
$$\bigl(xf'(x)\bigr)'=0$$
$$xf'(x)=C_{1}$$
$$f(x)=C_{1}\ln x+C_{2}$$
The verification:
$$f(f’(x))=C_1\ln\left(\frac {C_1}x\right)+C_2=-(C_1\ln (x)+C_2).$$
So $C_2=-\frac{C_1\ln C_1}2$ and 
$$f(x)=\frac{\ln Kx}{K^2}.$$
PS. In general, we cannot a priori assume that $f$ is twice differentiable. For instance, consider a similar situation. Let $g:\Bbb R\to\Bbb R$ be a function such that $g(x)=x^2$ if  $x\ge 0$ and $g(x)=-x^2$ if  $x< 0$. Then $g’(x)=|x|$, $g(g’(x))=x^2$ is a differentiable function, but $g’’(0)$ does not exist. So a question remains open for a not twice differentiable function $f$. 
