If $$f(x) = log frac{(1+x)}{(1-x)}$$ then $$f(y) = log frac{(1+y)}{(1-y)}$$ Also Log (A*B)= Log A + Log B  f(x)+f(y) = $$ log frac{(1+x)(1+y)}{(1-x)(1-y)}$$ solving we get $$log { frac{1+ frac{(x+y)}{(1+xy)}}{1- frac{(x+y)}{(1+xy)}}}$$ Hence option B.