(IMO SL 2018 A1)(M-D) Determine the functions $f: \mathbb{Q}_{+}^{*} \rightarrow \mathbb{Q}_{+}^{*}$ such that

$$
f\left(x^{2} f(y)^{2}\right)=f(x)^{2} f(y)
$$
$c=f(1)$ fixed and we set $y=1$ and then for all $x$ we have

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

and with $x=1$ we have $f\left(c^{2}\right)=c^{2}$. Then we set $x=y=1$ and we have $f\left(c^{2}\right)=c^{3}$ and thus $c=1$.

From there, $f\left(x^{2}\right)=f\left(x^{2}\right)$ and with $x=1$

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

We fix an arbitrary $y$ and consider $l=f(y)$. We then write $f\left((y x)^{2}\right)=l f(x)^{2}=f(y x)^{2}$ and then write that $f(x)^{2}=\left(f^{k}(x)\right)^{2 k}$ and thus $l^{\frac{1}{2 k}}=\frac{f^{k}(x)}{f^{k}(y x)} \in \mathbb{Q}$ and thus $l=1$. Finally, we have $f \equiv 1$.