Find the number of ordered pairs (m,n) of positive integers satisfying 1/m+1/n+1/mn=1/6.

Multiply by $6mn$: $6n+6m+6=mn$, so $mn-6m-6n=6$. Adding $36$ to both sides gives $(m-6)(n-6)=42$. Since $m,n\ge1$, each of $m-6,\ n-6$ is at least $-5$. If both factors were negative, each would lie in $\{-5,\dots,-1\}$ and their product would be at most $25<42$, so both are positive divisors of $42$ with product $42$. The positive divisors of $42=2\cdot3\cdot7$ number $2^3=8$, giving the ordered pairs $(m-6,n-6)=(1,42),(2,21),(3,14),(6,7)$ and their reversals, i.e. $(7,48),(8,27),(9,20),(12,13)$ and $(48,7),(27,8),(20,9),(13,12)$. A brute-force search over $m,n<2000$ confirms exactly $8$ pairs. Note that $m=6$ or $n=6$ is impossible, since it would make $(m-6)(n-6)=0\ne42$.