Let $P(x,y)$ be the assertion. $P(1,1)$. $f(1) = 1$ $P(x,1)$. $f(x) \cdot \gcd (f(x), f(\frac{1}{x}))= x \cdot f(\frac{1}{x}) \to f(x)^2 = x \cdot \text{lcm} (f(x), f(\frac{1}{x}))$. We lcm denotes the lowest common multiple. From that, using the similiar result $P(\frac{1}{x},1)$, we have $f(x) = x \cdot f(\frac{1}{x})$ and $\gcd (f(x), f(\frac{1}{x})) = 1$. But $\gcd (f(x), f(\frac{1}{x})) = \gcd(x \cdot f(\frac{1}{x}), f(\frac{1}{x})) = \gcd(\{x\} \cdot f(\frac{1}{x}), f(\frac{1}{x}))$. Where $\{x\}$ denotes the fractional part. If $\{x\} = 0$, i.e., $x$ is an integer, we have $f(\frac{1}{x}) = 1$. Now, let $\{x\} = \frac{a}{b}$, with, $\gcd(a, b) = 1$, then $b \mid f(\frac{1}{x})$ and let $c = \frac{f(\frac{1}{x})}{b}$, we have $\gcd (f(x), f(\frac{1}{x})) = \gcd(a \cdot c, b \cdot c) = c = \frac{f(\frac{1}{x})}{b}$. So we have $f(x) = b \cdot x$, where $b$ is the least integer such that $b \cdot x$ is an integer, i.e. the denominator of the reduced fraction. Meaning, $f(x)$ is the numerator of $x$. Q.E.D.

socrates wrote: Find all functions $f : \Bbb{Q}_{>0}\to \Bbb{Z}_{>0}$ such that $$f(xy)\cdot \gcd\left( f(x)f(y), f(\frac{1}{x})f(\frac{1}{y})\right) = xyf(\frac{1}{x})f(\frac{1}{y}),$$for all $x, y \in \Bbb{Q}_{>0,}$ where $\gcd(a, b)$ denotes the greatest common divisor of $a$ and $b.$ Let $P(x, y)$ denote the assertion given in the problem statement. $P(1,1) \Longleftrightarrow f(1)=1$ $P(x,1) \Longleftrightarrow f(x) \gcd \left (f(x),f \left( \frac {1} {x}\right ) \right)=x f \left(\frac {1} {x} \right) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (1)$ $P(1,x) \Longleftrightarrow f \left(\frac{1}{x}\right)\gcd\left (f(x),f \left(\frac{1}{x}\right)\right)=\frac{1}{x} f(x) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (2)$ Combining $ (1) \;\;$ and $\;(2)$ $\Longleftrightarrow f(x)=xf \left( \frac{1}{x}\right ) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (3)$ Putting $(3) $ on the original equation it becomes $ f(xy)\gcd \left(xyf \left(\frac {1} {x} \right)f \left(\frac {1} {x} \right),f\left( \frac {1} {x} \right)f \left(\frac {1} {x} \right) \right)=xyf \left(\frac {1} {x} \right)f \left(\frac {1} {y} \right)$ If $ x,y$ are both integers then $f(xy)=xy \Longleftrightarrow f(x)=x$ In $(3) $ we put $x$ integer $ \Longleftrightarrow f \left(\frac{1}{x}\right)=1$ whenever $x$ is integer Now , $P \left(x,\frac{1}{y} \right ) $ where $x,y $ are integers such that $\gcd(x,y)=1$ we get the solution $f \left(\frac{x}{y} \right )=x $

The answer is $f(x)$ equals the numerator of $x$ when written in simplest form. It is easy to check that this works. We now prove this is the only solution. Let $P(x,y)$ denote the assertion. Then $P(1,1)$ readily gives $f(1)=1$, and $P(x,1)$ gives the statement \[f(x)\gcd\left(f(x),f\left(\frac1x\right)\right)=xf\left(\frac1x\right).\tag{$\star$}\]Considering $(\star)$ with $x$ replaced by $1/x$, we have \[\frac{xf(1/x)}{f(x)}=\gcd\left(f(x),f\left(\frac1x\right)\right)=\frac{f(x)}{xf(1/x)}\implies x=\frac{f(x)}{f(1/x)}.\]Substituting $f(x)=xf(1/x)$ into $(\star)$, we have \[\gcd\left(f(x),f\left(\frac1x\right)\right)=1.\]Each positive rational $x$ may be uniquely expressed as the quotient of two coprime positive integers, so $f(x)$ is the numerator and $f(1/x)$ is the denominator. This completes the proof.

Very unique. Denote the assertion by $P(x,y).$ We note that $f(1)=1.$ Dividing $P(x,1)$ by $P(x^{-1},1)$ implies $f(x)=xf(x^{-1}).$ Then reapplication of $P(x,1)$ gives $\gcd (xf(x^{-1}), f(x^{-1}))=1$ and hence $f(x^{-1})=1$ and consequently $f(x)=x$ for integers. Finally $P(u,v^{-1})$ implies $f(uv^{-1})=u$ if $\gcd (u,v)=1.$ This function clearly satisfies