If f (x_1)-f (x_2)=f( x_1-x_2 1-x_1 x_2 ) for x_1, x_2 ∈(-1,1) ,...

$$\begin{aligned} & f\left(x_1\right)-f\left(x_2\right)=f\left(\frac{x_1-x_2}{1-x_1 x_2}\right) \\ & x_1, x_2 \in(-1,1) \end{aligned}$$$then$f(x)=\log \frac{(1-x)}{(1+x)} \begin{gathered} f\left(x_1\right)=\log \frac{1-x_1}{1+x_1} \quad f\left(x_2\right)=\log \frac{1-x_2}{1+x_2} \\ $$\begin{aligned} f\left(x_1\right)-f\left(x_2\right) & =\log \frac{1-x_1}{1+x_1}-\log \frac{1-x_2}{1+x_2} \\ & =\log \frac{\left(1-x_1\right)}{\left(1+x_1\right)} \times \frac{\left(1+x_2\right)}{\left(1-x_2\right)} \\ = & \log \frac{\left(1-x_1+x_2-x_1 x_2\right)}{\left(1+x_1-x_2-x_1 x_2\right)} \\ = & \log \frac{\left(1-x_1 x_2\right)-\left(x_1-x_2\right)}{\left(1-x_1 x_2\right)+\left(x_1-x_2\right)} \end{aligned}$$\end{gathered}=\log \frac{1-\left(\frac{x_1-x_2}{1-x_1 x_2}\right)}{1+\left(\frac{x_1-x_2}{1-x_1 x_2}\right)}f\left(x_1\right)-f\left(x_2\right)=f\left(\frac{x_1-x_2}{1-x_1 x_2}\right)$