Logarithmic Identity

$f(x) + f(y) = \log\left(\frac{1+x}{1-x}\right) + \log\left(\frac{1+y}{1-y}\right) = \log\left(\frac{(1+x)(1+y)}{(1-x)(1-y)}\right) = \log\left(\frac{1+x+y+xy}{1-x-y+xy}\right)$. Notice that $f\left(\frac{x+y}{1+xy}\right) = \log\left(\frac{1 + \frac{x+y}{1+xy}}{1 - \frac{x+y}{1+xy}}\right) = \log\left(\frac{1+xy+x+y}{1+xy-x-y}\right)$. Hence, $f(x) + f(y) = f\left(\frac{x+y}{1+xy}\right)$.