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 $b.$
1. **Define the assertion**: Let \( P(x, y) \) be the assertion of the given functional equation:
   \[
   f(xy) \cdot \gcd\left( f(x)f(y), f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right) \right) = xy f\left(\frac{1}{x}\right) f\left(\frac{1}{y}\right)
   \]
   for all \( x, y \in \mathbb{Q}_{>0} \).

2. **Evaluate \( P(1, 1) \)**: 
   \[
   P(1, 1): f(1 \cdot 1) \cdot \gcd\left( f(1)f(1), f\left(\frac{1}{1}\right)f\left(\frac{1}{1}\right) \right) = 1 \cdot 1 \cdot f\left(\frac{1}{1}\right) f\left(\frac{1}{1}\right)
   \]
   Simplifying, we get:
   \[
   f(1) \cdot \gcd(f(1)^2, f(1)^2) = f(1)^2
   \]
   Since \(\gcd(f(1)^2, f(1)^2) = f(1)^2\), we have:
   \[
   f(1) \cdot f(1)^2 = f(1)^2 \implies f(1) = 1
   \]

3. **Evaluate \( P(x, 1) \)**:
   \[
   P(x, 1): f(x \cdot 1) \cdot \gcd\left( f(x)f(1), f\left(\frac{1}{x}\right)f\left(\frac{1}{1}\right) \right) = x \cdot 1 \cdot f\left(\frac{1}{x}\right) f(1)
   \]
   Simplifying, we get:
   \[
   f(x) \cdot \gcd(f(x), f\left(\frac{1}{x}\right)) = x f\left(\frac{1}{x}\right)
   \]
   Let \( g(x) = f\left(\frac{1}{x}\right) \). Then the equation becomes:
   \[
   f(x) \cdot \gcd(f(x), g(x)) = x g(x)
   \]

4. **Evaluate \( P\left(\frac{1}{x}, 1\right) \)**:
   \[
   P\left(\frac{1}{x}, 1\right): f\left(\frac{1}{x} \cdot 1\right) \cdot \gcd\left( f\left(\frac{1}{x}\right)f(1), f\left(\frac{1}{\frac{1}{x}}\right)f\left(\frac{1}{1}\right) \right) = \frac{1}{x} \cdot 1 \cdot f(x) f(1)
   \]
   Simplifying, we get:
   \[
   g(x) \cdot \gcd(g(x), f(x)) = \frac{1}{x} f(x)
   \]
   Since \(\gcd(f(x), g(x)) = \gcd(g(x), f(x))\), we have:
   \[
   g(x) \cdot \gcd(g(x), f(x)) = \frac{1}{x} f(x)
   \]

5. **Combine the results**:
   From the equations:
   \[
   f(x) \cdot \gcd(f(x), g(x)) = x g(x)
   \]
   \[
   g(x) \cdot \gcd(g(x), f(x)) = \frac{1}{x} f(x)
   \]
   We can infer that \( f(x) = x g(x) \) and \( g(x) = \frac{f(x)}{x} \). Therefore:
   \[
   f(x) = x f\left(\frac{1}{x}\right)
   \]

6. **Determine \( f(x) \)**:
   Let \( x = \frac{a}{b} \) where \( \gcd(a, b) = 1 \). Then:
   \[
   f\left(\frac{a}{b}\right) = \frac{a}{b} f\left(\frac{b}{a}\right)
   \]
   Since \( f\left(\frac{b}{a}\right) \) must be an integer, let \( f\left(\frac{b}{a}\right) = k \). Then:
   \[
   f\left(\frac{a}{b}\right) = \frac{a}{b} k
   \]
   For \( f\left(\frac{a}{b}\right) \) to be an integer, \( k \) must be \( b \). Thus:
   \[
   f\left(\frac{a}{b}\right) = a
   \]
   Therefore, \( f(x) \) is the numerator of \( x \) when \( x \) is expressed in its lowest terms.

The final answer is \( \boxed{ f(x) = \text{numerator of } x } \).