functional equation of type $f(x)-f(y) = f\left(\frac{x-y}{1-xy}\right)$ If $f(x)-f(y) = f\left(\frac{x-y}{1-xy}\right)$ and $f$ has domain equal to $(-1,1)$, then which of the following is the function satisfying the given functional equation?
Options:
(a) $2-\ln\left(\frac{1+x}{1-x}\right)$ 
(b) $\ln\left(\frac{1-x}{1+x}\right)$
(c) $\frac{2x}{1-x^2}$
(d) $\tan^{-1}\left(\frac{1+x}{1-x}\right)$
From the options given we can easily get $f(x) = \ln\frac{1-x}{1+x}$,
but how can I solve using analytical method?  Help me please.

Taking $y=0$ gives $f(0) = 0$. Using this we can rewrite the functional equation as
$$\frac{f(x)-f(y)}{x-y} = \frac{1}{1-xy}\frac{f\left(\frac{x-y}{1-xy}\right) - f(0)}{\left(\frac{x-y}{1-xy}\right)}$$
Now assuming $f$ is differentiable at $x=0$ we can take the limit $y\to x$ above to obtain the ODE $$f'(x) = \frac{1}{1-x^2}f'(0) = \frac{f'(0)}{2}\left[\frac{1}{1-x} + \frac{1}{1+x}\right]$$
which can be integrated with the initial condition $f(0) = 0$ to get the general solution $f(x) = C\log\left(\frac{1-x}{1+x}\right)$ where $C = -\frac{f'(0)}{2}$ is a free constant.
