First, let's ignore the requirement that the product can't be $ 6 $. Then I can pick the first blue face in $ 6 $ ways, and the second blue face in $ 5 $ ways, making $ 6\cdot 5 = 30 $ choices in all. But we've actually counted each possible result twice, because it makes no difference which of the two blue faces I chose first and which I chose second. So the number of different pairs of faces is really $ (6\cdot 5)/2 $, or $ 15 $. Now we exclude the pairs which have a product of $ 6 $. There are two such pairs: $ \{1,6\} $ and $ \{2,3\} $. That leaves me $ \boxed{13} $ pairs of faces I can paint blue.