What is the derivative of $f(x)$ if $f(x)+f(y)=f\left(\frac{x+y}{1-xy}\right)$?

Ok so I have almost solved it.

$f'(x)=f'\left(\frac{x+y}{1-xy}\right)\left(\frac{(1-xy)-(x+y)(-y)}{(1-xy)^2}\right)$

Let $x=0$, then $f'(0)=f'(y)(1+y^2)$, thus $f'(y)=\frac{c}{1+y^2}$ for all $y$ in $\mathbb{R}$.

So we have $f(y)=c*\arctan y+C$ where $C=0$ since $f(0)=0$.

I know the function that I've gotten satisfies the functional equation if $xy<1$.

So if I want to show that the functional equation is satisfied, must I prove that $\arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right)$ for $xy<1$ or is it enough that I got $f(y)$ based on $x=0$ and $y$ is any real number so $xy<1$?