Let f(x)=(αx)/(x+1),x≠-1. If f(f(x))=x, then value of α is

$f(f(x))=\frac{αf(x)}{f(x)+1}=\frac{α^2x}{(α+1)x+1}$

Thus $f ( f (x)) = x⇔α^2x = (α + 1)x^2 + x$

$⇔(α^2-1)x = (α + 1)x^2$

$⇔(α+1)((α - 1)-x)=0$

This equation is true for all x, if $α = –1$