Find $f(x)$ such that $f(x)^{2} f\left(\frac{1-x}{1+x}\right)=64 x$ for $x$ not $0, \pm 1$.
Put $x=(1-y) /(1+y)$, then $(1--x) /(1+x)=y$, so $f((1-y) /(1+y))^{2} f(y)=64(1-y) /(1+y)$. Hence $f((1-x) /(1+x))^{2} f(x)=64(1-x) /(1+x)$. But $f(x)^{4} f((1-x) /(1+x))^{2}=64^{2} x^{2}$, so $f(x)^{3}=64 x^{2}(1+x) /(1-x)$. Hence $f(x)=4\left(x^{2}(1+x) /(1-x)\right)^{1 / 3}$.