$F({x+y\over 1+xy}) = F(x)\cdot F(y)$ for all real $x, y$ except $-1$

Question

If $F(x)$ is a differentiable function that satisfies the above functional equation for all real x and y except -1. If $F(0) \ne 0$ and $F'(0) = 1$ then $F(x) = ? $

Answer 1

Hint

Let $x=\tan x_1$ and $y=\tan y_1$ with $|x_1|<{\pi\over 2}$ and $|y_1|<{\pi\over 2}$. Then by defining $$g(x)=e^{f(\tan x)}$$ we have $$g(x_1+y_1)=g(x_1)+g(y_1)$$