There is no function $f\colon Q\to Q$ such that $$f(xf(y))=\frac{f(f(x))}y \tag{1}$$ for all $x$ and $y$ (in $Q$), where $Q:=\mathbb Q_{+}^{*}$.
Indeed, for $x=1$ equality (1) is $$f(f(y))=\frac{f(b)}y,$$ where $b:=f(1)$. Replacing here $y$ by $x$, from (1) we get $$f(xf(y))=\frac{f(b)}{xy}.$$ This with $y=1$ yields $$f(xb)=\frac{f(b)}{x},$$ or $$f(z)=\frac cz$$ for $c:=bf(b)$ and all $z\in Q$. Now (1) becomes $y/x=x/y$ for all $x,y$ in $Q$, which is clearly false.