SOLUTION: (6xy-4x-9y+6)/(6y^2-13y+6)

 6xy-4x-9y+6 
  6y²-13y+6

First we must factor the numerator:

6xy - 4x - 9y + 6

Out of the first two terms we factor out 2x:

2x(3y - 2) - 9y + 6

Out of the last two terms we factor out -3

2x(3y - 2) - 3(3y - 2)

Now we notice that there is a common factor
of (3y - 2).  So we factor that red factor
out putting the black parts in parentheses:

(3y - 2)(2x - 3)

or writing it all in black:

(3y - 2)(2x - 3)

Next we must factor the denominator:

6y² - 13y + 6

Multiply the red 6 by the purple 6, getting 36

Think of two integers which have product 36 and SUM the green 13.
(Note: the reason it's SUM and not DIFFERENCE is because the last
sign in the trinomial is +. If it had been - we would have said
"DIFFERENCE" here).

Anyway, two integers whose product is 36 and whose sum is 13 are
9 and 4.  So we rewrite 13 as either 4 + 9 or 9 + 4, whichever you
choose. I will choose 9 + 4:

So the denominator 

    6y² - 13y + 6

becomes

    6y² - (9 + 4)y + 6

Distribute to remove the parentheses:

    6y² - 9y - 4y + 6

Factor 3y out of the first two terms:

   3y(2y - 3) - 4y + 6

Factor -2 out of the last two terms:

   3y(2y - 3) - 2(2y - 3)

Now we notice that there is a common factor
of (2y - 3).  So we factor that red factor
out putting the black parts in parentheses:

(2y - 3)(3y - 2)

or writing it all in black:

(2y - 3)(3y - 2)

Now putting the factored numerator over the factored denominator:

 (3y - 2)(2x - 3) 
 (2y - 3)(3y - 2)

Now we cancel the (3y - 2)'s

     1
 (3y - 2)(2x - 3) 
 (2y - 3)(3y - 2)
             1

and all that's left is 

 (2x - 3)
 (2y - 3)

We do not need the parentheses:

 2x - 3 
 2y - 3  

Caution: do not try to cancel anything else
because neither the 2's not the -3's are
factors of the numerator and denominator. 

Edwin