Example 10 (IMO 2008, Problem 4) Find all functions $f:(0,+\infty) \rightarrow(0,+\infty)$ satisfying that for all positive real numbers $\omega, x, y, z, \omega x=y z$ we have: $\frac{(f(\omega))^{2}+(f(x))^{2}}{f\left(y^{2}\right)+f\left(z^{2}\right)}$ $=\frac{\omega^{2}+x^{2}}{y^{2}+z^{2}}$.
Let $\omega=x=y=z=1$, we get: $(f(1))^{2}=f(1)$ $\Rightarrow f(1)=1$.

For any $t>0$, let $\omega=t, x=1, y=z=\sqrt{t}$, we get: $\frac{(f(t))^{2}+1}{2 f(t)}=\frac{t^{2}+1}{2 t}$,

Eliminating the denominator and rearranging: $(t f(t)-1)(f(t)-t)=0$, so for each $t>0$ we have $f(t)=t$ or $f(t)=\frac{1}{t}$.

If there exist $b, c \in(0,+\infty)$, such that $f(b) \neq b$, $f(c) \neq \frac{1}{c}$, then by (1) we know that $b, c$ are not equal to 1, and $f(b)=\frac{1}{b}, f(c)=c$. Let $\omega=b, x=c, y=z=\sqrt{b c}$, then $\frac{\frac{1}{b^{2}}+c^{2}}{2 f(b c)}=\frac{b^{2}+c^{2}}{2 b c}$, so $f(b c)=\frac{c+b^{2} c^{3}}{b\left(b^{2}+c^{2}\right)}$. Since $f(b c)=b c$ or $f(b c)=\frac{1}{b c}$, if $f(b c)=b c$, then $b c=\frac{c+b^{2} c^{3}}{b\left(b^{2}+c^{2}\right)} \Rightarrow b^{4} c=c \Rightarrow b=1$ contradiction; if $f(b c)=\frac{1}{b c}$, then $\frac{1}{b c}=\frac{c+b^{2} c^{3}}{b\left(b^{2}+c^{2}\right)} \Rightarrow b^{2} c^{4}=b^{2} \Rightarrow c=1$ contradiction.

Therefore, $f(x)=x(x \in(0,+\infty))$ or $f(x)=\frac{1}{x}(x \in(0,+\infty))$ which, upon verification, satisfy the conditions.