First let's prove that $f$ is injective: suppose that $f(a) = f(b)$. Then
$$\begin{aligned} \frac{2a f(f(a))}{2b f(f(b))} &= \frac{f(a)(a+f(f(a)))}{f(b)(b+f(f(b)))} \\ &\iff \frac{a}{b} = \frac{a+f(f(a))}{b+f(f(b))} = \frac{f(f(a))}{f(f(b))} = 1 \\ &\iff a = b. \end{aligned}$$

Then $f$ is invertible. Since the given equation is equivalent to
$$\frac{2}{f(x)} = \frac{1}{x} + \frac{1}{f(f(x))},$$
defining $\frac{1}{f^{(n)}(x)} = a_n,\ n \in \mathbb{Z}$, we have, plugging $x \to f^{(n)}(x)$,
$$2a_{n+1} = a_n + a_{n+2} \iff a_{n+2} - a_{n+1} = a_{n+1} - a_n,$$
that is, $a_n$ is an arithmetic progression. Since $a_n > 0$ for all $n, x$, it has to be constant, and $a_1 = a_0 \iff \frac{1}{f(x)} = \frac{1}{x} \iff f(x) = x$, which can be easily verified to satisfy the equation.