6. find all functions $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$so that for all positive rational numbers $x, y$ holds

$$
f\left(f(x)^{2} y\right)=x^{3} f(x y)
$$
Solution With $y=1$ follows

$$
f\left(f(x)^{2}\right)=x^{3} f(x)
$$

in particular, $f$ is injective. If you replace $x$ with $x y$, you get

$$
f\left(f(x y)^{2}\right)=x^{3} y^{3} f(x y)
$$

On the other hand, you can replace $y$ in the original equation with $f(y)^{2}$ and, using this equation again, obtain

$$
f\left(f(x)^{2} f(y)^{2}\right)=x^{3} f\left(f(y)^{2} x\right)=x^{3} y^{3} f(x y)
$$

A comparison of the last two equations together with the injectivity of $f$ finally yields

$$
f(x y)=f(x) f(y)
$$

Due to the multiplicative nature of $f$, the following applies in particular

$$
f\left(x^{m}\right)=f(x)^{m}
$$

for every integer $m$ and thus also for every rational number $m$, provided that the expressions involved are in $\mathbb{Q}^{+}$. In particular, (1) becomes

$$
f(f(x))=\sqrt{x^{3} f(x)}
$$

Furthermore, you now get

$$
f(x f(x))=f(x) f(f(x))=f(x) \sqrt{x^{3} f(x)}=(x f(x))^{3 / 2}
$$

and with (2) and complete induction

$$
f^{n}(x f(x))=(x f(x))^{3^{n} / 2^{n}}
$$

for all natural numbers $n$. Here, the right-hand side is a rational number for all $n$, but this is only possible for $x f(x)=1$. Therefore, $f(x)=\frac{1}{x}$ applies for $x \in \mathbb{Q}^{+}$. Obviously, this is really a solution to the original equation.