Find all functions that satisfy $f(x^2f(y)^2)=f(x)^2f(y)$.

Find all $f:\mathbb Q_{>0}\to \mathbb Q_{>0}$ such that $$f(x^2f(y)^2)=f(x)^2f(y)$$ This is from the IMO Shortlist 2018, and I just want to know if my solution is valid, here is the official solution

set $x=1/f(y)^2$ we get $$f\left(\frac{1}{f(y)^2}\right)= f\left(\frac{1}{f(y)^2}\right)^2 f(y)$$

We can cancel these two because of the domain and the range of the $f$, which means $$f \left(\frac{1}{f(y)^2} \right)= \frac{1}{f(y)} $$ $$f(t)=\sqrt{t}$$ but that doesn’t satisfy the range of the function and it’s not even a solution to the functional equation.

(Not an answer) To elaborate on my comment, the mistake OP made in

$$f \left(\frac{1}{f(y)^2} \right)= \frac{1}{f(y)} \Rightarrow f(t)=\sqrt{t} \, \forall t $$

is that the substitution of $ t = \frac{1}{f(y)^2}$ requires there to be a $y$ for that particular value of $t$.

EG If there are no solutions to $ f(y) ^ 2 = 4$, then we cannot conclude that $f( 0.25) = f( \frac{1}{4} ) = \frac{1}{2} $.

Thus, all that we have is

If there is some $y$ such that $ t = \frac{1}{f(y)^2}$, then $f(t) = \sqrt{t}$.

Assuming that we can show $f(y) = 1$ for some $y$ (which is a big assumption, since the official solution doesn't seem to establish specific values), then we can conclude via this approach that indeed $f(1) = 1$.