Find the number of pairs of positive integers $(x,y)$ that satisfy the equation:

Step 1: Clear the denominators by multiplying the entire equation by $6xy$:

Step 2: Use Simon's Favorite Factoring Trick by adding $36$ to both sides to make it factorable:

Step 3: Since $x$ and $y$ must be positive integers, $x-6$ and $y-6$ must be integer factors of $36$. Let $A=x-6$ and $B=y-6$. Then $A\cdot B=36$.

• Since $x,y>0$, we must have $x-6>-6$ and $y-6>-6$.

• If both $A$ and $B$ were negative (e.g. $A=-6,B=-6$), we would get $x=0$ (invalid as $x$ must be positive). If $A<-6$, then $x<0$, which is also invalid. Therefore, both $A$ and $B$ must be positive integers.

Step 4: The number of pairs of positive integers $(x,y)$ is exactly the number of positive divisors of $36$.
The prime factorization of $36$ is:

The number of divisors is:

Each divisor $d$ corresponds to a unique pair $x-6=d⟹x=6+d$ and $y-6=36/d⟹y=6+36/d$. Both will be positive integers.

Thus, there are exactly $9$ pairs.