can there be a function $f:\mathbb Q_{+}^{*}\longmapsto\mathbb Q_{+}^{*}$ such that $f(xf(y))=\frac{f(f(x))}{y}$?

Question

Problem: Can an $f$ function be created where:$$f\colon\mathbb Q_{+}^{*}\to \mathbb Q_{+}^{*}$$ The function is defined on the set of fully positive rational numbers and is achieved: $\forall(x,y)\in \mathbb Q_{+}^{*}\times\mathbb Q_{+}^{*},f(xf(y))=\frac{f(f(x))}{y}$

This question is similar to one of the Olympiad questions that I was very passionate about and used several ideas to solve this problem, but I did not arrive at any result from one of them by using the basic theorem in arithmetic that states that there is a corresponding application between $(\mathbb Q_{+}^{*})$and $(\mathbb Z^{\mathbb N})$ where: $$\left\{\mathbb Z^{\mathbb N} =\text{ A set of stable sequences whose values ​​are set in} \quad\mathbb Z\right\}$$ This app is defined like this $$\varphi\colon\mathbb Z^{\mathbb N}\to \mathbb Q_{+}^{*} ,(\alpha_n)_{n\in\mathbb N}\longmapsto \prod_{n\in\mathbb N} P_n^{\alpha_n}$$ Where:$$\mathbb P=\left\{P_k:k\in\mathbb N\right\}\text{ is the set of prime numbers} $$ And put $x=\prod_{n\in\mathbb N}P_n^{\alpha_n},\quad y=\prod_{n\in\mathbb N }P_n^{\beta_n},\text{and}\quad $ $$f(\prod_{n\in\mathbb N}P_n^{\alpha_n})=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}}\right)$$

. \begin{align*} xf(y)&=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n}^{\beta_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\beta_{2n}}\right)\\ &=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n}+\beta_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{\alpha_{2n+1}-\beta_{2n}}\right)\\ \end{align*}

$\implies$ \begin{align*} f(xf(y))&=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}-\beta_{2n}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}-\beta_{2n+1}}\right)\\ &=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}}\right)\left(\prod_{n\in\mathbb N}P_{2n}^{-\beta_{2n}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\beta_{2n+1}}\right)\\ &=\frac{\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}}\right)}{\left(\prod_{n\in\mathbb N}P_{n}^{\beta_{n}}\right)}\\ &=\frac{f(x)}{y}\\ \end{align*}

However, this did not help me create this method

I need an idea or suggestion to solve this problem if possible and thank you for your help

Note: $(\alpha_n)_{n\in\mathbb N}\quad \text{is a stable sequence}\leftrightarrow \forall n\in\mathbb N ,\exists n_0\in\mathbb N :\left( n\geq n_0 \quad \alpha_{n}=0\right) $

Answer 1

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.