$f(x) = 1$ for all $x \in Q_{>0}$.
Take any $a, b \in Q_{>0}$. By substituting $x = f(a)$, $y = b$ and $x = f(b)$, $y = a$ into the assumption of the problem we get
$$(f(f(a)))^2 f(b) = f(f(a)^2 f(b)^2) = f(f(b))^2 f(a),$$
which yields
$$\frac{f(f(a))^2}{f(a)} = \frac{f(f(b))^2}{f(b)} \text{ for all } a, b \in Q_{>0}.$$
In other words, this shows that there exists a constant $C \in Q_{>0}$ such that
$$f(f(a))^2 = C f(a),$$
$$\left(\frac{f(f(a))}{C}\right)^2 = \frac{f(a)}{C} \quad \text{for all } a \in \mathbb{Q}_{>0}. \qquad (1)$$
Denote by $f^n(x) = \underbrace{f(f(\dots(f(x))\dots))}_{n}$ the $n$th iteration of $f$. Equality (1) yields
$$\frac{f(a)}{C} = \left(\frac{f^2(a)}{C}\right)^2 = \left(\frac{f^3(a)}{C}\right)^4 = \dots = \left(\frac{f^{n+1}(a)}{C}\right)^{2^n}$$
for all positive integer $n$. So, $f(a)/C$ is the $2^n$-th power of a rational number for all positive integer $n$. This is impossible unless $f(a)/C = 1$,

since otherwise the exponent of some prime in the prime decomposition of $f(a)/C$ is not divisible by sufficiently large powers of 2. Therefore, $f(a) = C$ for all $a \in \mathbb{Q}_{>0}$.
Finally, after substituting $f \equiv C$ into the assumption of the problem we get $C = C^3$, whence $C = 1$. So $f(x) \equiv 1$ is the unique function satisfying the assumption of the problem.