There is no function $f\colon Q\to Q$ such 
$$f(xf(y))=\frac{f(f(x))}y \tag{1}$$
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 absurd.