Simplify [(6x^2*y^2)/(9x^2-9y^2)]/[(18x^2*y^3/(3x-3y)]

We notice that we have to calculate a division between 2
quotients. We'll have to multiply the first ratio with the inversed second ratio, where
the numerator becomes denominator and denominator becomes
numerator!

We'll re-write the denominator, that is a
difference of squares, as a product:

9x^2 - 9y^2 =
(3x-3y)(3x+3y)

The expression will
become:

[(6x^2*y^2)/(3x-3y)(3x+3y)]*[(3x-3y)/(18x^2*y^3)]

We'll
write the denominator 18x^2*y^3=3*6x^2*y^2*y

We can now
simplify like terms, which are (3x-3y),and [(6x)^2y^2].

The
result will
be:

[1/(3x+3y)]*(1/3y)

[(6x^2*y^2)/(9x^2-9y^2)]/[(18x^2*y^3/(3x-3y)]=1/9y(x+y)