Finding $f$ such that $f(x)+f(y)=f\left(\frac{x+y}{1-xy}\right)$

Question

Determine all functions $f$, wich are everywhere differentiable and satisfy

$$f(x)+f(y)=f\left(\frac{x+y}{1-xy}\right)$$

for all real $x$ and $y$ with $x.y \ne 1$.

PS.: The expression sugest some relation with $\tan(x)$ but I can't go further. Any hint?

Answer 1

With $g:=f\circ \tan$, we have $$g(x)+g(y)=g(x+y) $$ for all $x,y$ with $x+y\ne\frac\pi 2+2k\pi$. This quickly leads to $g(x)=cx$.