Take
$$w = x = y = z = 1,$$
then we get $(f(1))^2 = f(1)$, so $f(1) = 1$.
For any real number $t > 0$, let $w = t$, $x = 1$, $y = z = \sqrt{t}$,
we get
$$\frac{(f(t))^2 + 1}{2f(t)} = \frac{t^2 + 1}{2t},$$
which implies $(tf(t) - 1)(f(t) - t) = 0$.
So, for any $t > 0$,
$$f(t) = t \quad \text{or} \quad f(t) = \frac{1}{t}. \qquad \textcircled{1}$$
Suppose there exist $b, c \in (0, +\infty)$ such that $f(b) \neq b$, $f(c) \neq \frac{1}{c}$. By ①, we get $b, c$ different from 1 and $f(b) = \frac{1}{b}$, $f(c) = c$.
Take $w = b, x = c, y = z = \sqrt{bc}$, then
$$\frac{\frac{1}{b^2} + c^2}{2f(bc)} = \frac{b^2 + c^2}{2bc},$$
i.e. $f(bc) = \frac{c + b^2c^3}{b(b^2 + c^2)}$.
By ①, $f(bc) = bc$ or $f(bc) = \frac{1}{bc}$. If $f(bc) = bc$, then
$$bc = \frac{c + b^2c^3}{b(b^2 + c^2)},$$
which yields $b^4c = c, b = 1$. Contradiction!

If $f(bc) = \frac{1}{bc}$, then
$$\frac{1}{bc} = \frac{c + b^2 c^3}{b(b^2 + c^2)},$$
that yields $b^2c^4 = b^2$, $c = 1$. Contradiction!
Therefore, only two functions: $f(x) = x, x \in (0, +\infty)$ or $f(x) = \frac{1}{x}, x \in (0, +\infty)$. It is easy to verify that these two functions satisfy the given conditions.