Information to the problem : it is extremly difficult because there was a typing mistake , the problem didnt meant to be like that. Still, one Person in Germany was able to solve the problem in vaimo , he got 11. In imo 2019.

Well, in principle the problem is not very hard, it is just quite unusual for an olympiad problem. Let $I=\text{Im}(f)$ and $J=\{f(x^2): x \in \mathbb{Q}^+\}$. The assumption can be written as $f(q^2z^2)=zf(q^2)$ for $z \in I, q \in \mathbb{Q}^+$. Write $a=f(1)$. First step: $I=J$ and this set is a subgroup of $\mathbb{Q}^+$ (under multiplication). We trivially have $J \subset I$. Putting $q=1$ we have $f(z^2)=az$ and hence $aI \subset J$. Putting $q=\frac{1}{a}, z=a$ we have $f\left(\frac{1}{a^2}\right)=1$. Putting $q=\frac{1}{az}$ we have $f\left(\frac{1}{a^2z^2}\right)=\frac{1}{z}$ and hence $\frac{1}{z} \in I$ for $z \in I$. Hence exchanging $z$ for $\frac{1}{z}$ we get $f\left(\frac{z^2}{a^2}\right)=z$ for $z \in I$ and hence $J \subset I$ so that $I=J$. For $y,z \in I$ we can now write $y=f(q^2)$ since $y \in J$ and hence $f(q^2z^2)=yz$ so that $I$ is indeed a group (we already checked the inverses). Second step: Description of $f$. We desribe the set of all possible solutions: Let $I \subset \mathbb{Q}^+$ be an arbitrary subgroup and let $g:(\mathbb{Q}^+)^2/I^2 \to I$ be arbitrary (i.e. to each coset of the squares w.r.t. $I^2$ we associate an arbitrary value in $I$). Then we can extend $f$ uniquely to $(\mathbb{Q}^+)^2$ using the rule $f(q^2z^2)=zf(q^2)$. Moreover, we can choose $f$ arbitrarily on the non-squares. On the other hand, from the first step it is clear that every solution $f$ must be of this form.