By substituting $y=1$, we get
$$f\left(f(x)^{2}\right)=x^{3} f(x)$$
Then, whenever $f(x)=f(y)$, we have
$$x^{3}=\frac{f\left(f(x)^{2}\right)}{f(x)}=\frac{f\left(f(y)^{2}\right)}{f(y)}=y^{3}$$
which implies $x=y$, so the function $f$ is injective.

Now replace $x$ by $x y$ in the previous equation, and apply the original equation twice, the second time to $(y, f(x)^{2})$ instead of $(x, y)$:
$$f\left(f(x y)^{2}\right)=(x y)^{3} f(x y)=y^{3} f\left(f(x)^{2} y\right)=f\left(f(x)^{2} f(y)^{2}\right)$$
Since $f$ is injective, we get
$$\begin{aligned} f(x y)^{2} & =f(x)^{2} f(y)^{2} \\ f(x y) & =f(x) f(y) \end{aligned}$$
Therefore, $f$ is multiplicative. This also implies $f(1)=1$ and $f\left(x^{n}\right)=f(x)^{n}$ for all integers $n$.

Then the functional equation can be re-written as
$$\begin{align*} f(f(x))^{2} f(y) & =x^{3} f(x) f(y), \\ f(f(x)) & =\sqrt{x^{3} f(x)} \end{align*}$$
Let $g(x)=x f(x)$. Then, we have
$$\begin{aligned} g(g(x)) & =g(x f(x))=x f(x) \cdot f(x f(x))=x f(x)^{2} f(f(x))= \\ & =x f(x)^{2} \sqrt{x^{3} f(x)}=(x f(x))^{5 / 2}=(g(x))^{5 / 2} \end{aligned}$$
and, by induction,
$$\underbrace{g(g(\ldots g}_{n+1}(x) \ldots))=(g(x))^{(5 / 2)^{n}}$$
for every positive integer $n$.

Consider this for a fixed $x$. The left-hand side is always rational, so $(g(x))^{(5 / 2)^{n}}$ must be rational for every $n$. We show that this is possible only if $g(x)=1$. Suppose that $g(x) \neq 1$, and let the prime factorization of $g(x)$ be $g(x)=p_{1}^{\alpha_{1}} \ldots p_{k}^{\alpha_{k}}$ where $p_{1}, \ldots, p_{k}$ are distinct primes and $\alpha_{1}, \ldots, \alpha_{k}$ are nonzero integers. Then the unique prime factorization is
$$\underbrace{g(g(\ldots g}_{n+1}(x) \ldots))=(g(x))^{(5 / 2)^{n}}=p_{1}^{(5 / 2)^{n} \alpha_{1}} \ldots p_{k}^{(5 / 2)^{n} \alpha_{k}}$$
where the exponents should be integers. But this is not true for large values of $n$, for example $\left(\frac{5}{2}\right)^{n} \alpha_{1}$ cannot be an integer when $2^{n} \nmid \alpha_{1}$. Therefore, $g(x) \neq 1$ is impossible.

Hence, $g(x)=1$ and thus $f(x)=\frac{1}{x}$ for all $x$.

The function $f(x)=\frac{1}{x}$ satisfies the equation:
$$f\left(f(x)^{2} y\right)=\frac{1}{f(x)^{2} y}=\frac{1}{\left(\frac{1}{x}\right)^{2} y}=\frac{x^{3}}{x y}=x^{3} f(x y)$$