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.
 A: 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.
A: Let's first change $y \mapsto-y$. Now, the functional equation in the question would be $$f(x)-f(-y)=f\left(\frac{x+y}{1+xy}\right)$$
Also by setting $x=0$ it turns out that $f$ is an odd function, i.e., $f(y)=-f(-y)$. Applying this to the previous equation gets $$f(x)+f(y)=f\left(\frac{x+y}{1+xy}\right)\label{eq1}\tag{1}$$
One way to find $f$ is to note the similarity between the expression $\frac{x+y}{1+xy}$ in equation (\ref{eq1}) and the following property of $\tanh$ function $$\tanh(a+b)=\frac{\tanh{a}+\tanh{b}}{1+\tanh{a}\tanh{b}}$$ By setting $x=\tanh{a}$ and $b=\tanh{b}$, the equation becomes $$\tanh(a+b)=\frac{x+y}{1+xy}$$ Therefore equation (\ref{eq1}) can be rewritten as
$$f(\tanh{a})+f(\tanh{b})=f(\tanh{(a+b)})$$
Obviously $f(u)=\tanh^{-1}{u}$ is a solution; since it cancels out all $\tanh(\cdot)$ functions and then the true statement $a+b=a+b$ will emerge. By using the definition of $\tanh u$ and a few manipulation, you can verify that $$\tanh^{-1}{u}=\frac12\log{\frac{1+u}{1-u}}$$
Moreover, due to the linearity, the function $f(u)=k\tanh^{-1}{u}$, where $k$ is a constant, is a solution either. By setting $k=-2$, you'll find (b) as the correct answer.
