Find all functions $ f: (0, \infty) \mapsto (0, \infty)$ (so $ f$ is a function from the positive real numbers) such that $$ \frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}$$ for all positive real numbers $ w,x,y,z,$ satisfying $ wx = yz.$
Attemp: Since the ratio applies to every quartet $ (w, x, y, z) $ such that $ wx = yz $ then we can choose $(w,x,y,z)=(1,1,1,1)$ and we have the following expression: $$\frac{(f(1))^2+(f(1))^2}{f(1^2)+f(1^2)}= \frac{1^2+1^2}{1^2+1^2}$$ $$f(1)=1$$
It seems valid to me but I can't prove it: Like $ (f (x)) ^ 2 = f (x ^ 2) $ and the function domain does not include zero, necessarily $ f (x) = x ^ n $. Now let's choose $ (w, x, y, z) = (1,1,2, \frac {1} {2}) $ and have the following expression: $$\frac{(f(1))^2+(f(1))^2}{f(2^2)+f((\frac{1}{2})^2)}= \frac{1^2+1^2}{2^2+(\frac{1}{2})^2}$$ $$4^n+4^{-n}= \frac{17}{4}$$ $$4^n= \frac{17 +-15}{8}$$
So $ n = 1 $ or $ n = -1 $. So the solutions are $ f (x) = x $ and $ f (x) = \frac {1} {x} $
Is my solution correct? If $ (f (x)) ^ 2 = f (x ^ 2) $ and $ f: (0, inf) \to (0, inf) $, you can prove that $ f (x) = x ^ n $ ?