If f(x)=(x-1)/(x+1), then f(2x) is equal to

$f(x)=\frac{x-1}{x+1}⇒f(2x)\frac{2x-1}{2x+1}$   …(i)

Also $xf (x)+ f (x) = x −1 ⇒x=\frac{f(x)+1}{1-f(x)}$   …(ii)

From equation (i) and (ii) 

$⇒f(2x)=\frac{2\left(\frac{f(x)+1}{1-f(x)}\right)-1}{2\left(\frac{f(x)+1}{1-f(x)}\right)+1}⇒f(2x)=\frac{3f(x)+1}{f(x)+3}$