This is an extension to : Functions $f: \Bbb Q_{+} \to \Bbb Z$ such that $f(\frac{1}{x}) = f(x)$ and $(x+1)f(x-1)= xf(x)$
What can be said about functions $f : \Bbb Q^*_+ \to \Bbb Q$ such as $f(\tfrac1x)=f(x)$ and $f(x) = (1+\tfrac1x)f(x-1)$ for all $x$?
(the difference with the original question being the functions' codomain).
Every strictly positive rational can be transformed into $1$ by applying $x \mapsto x-1$ and $x \mapsto 1/x$.
This is immediate by induction on $p+2q$ : If $p/q > 1$ we can transform it into $p/q-1 = (p-q)/q$, and if $p/q < 1$ we can transform it into $q/p$. At every step, $p+2q$ decrease.
As a result, for every rational $x$ there is a rational $y(x)$ such that for every solution $f$ of the problem, $f(x) = y(x) f(1)$.
$y(x)$ itself has to be the solution of the problem that has value $1$ at $1$, and we know from the previous question that $y(p/q) = (p+q)/2$.
Therefore, for any solution $f$, we have $f(p/q) = (p+q)f(1)/2$ for all coprime $p$ and $q$. And every choice of $f(1) \in \Bbb Q$ gives a solution.