Division by a monomial help..???

1)
(9xy^2 - 6y^3)/3x
= 3y^2(3x - 2y)/3x
= [3/3][y^2(3x - 2y)/x]
= y^2(3x - 2y)/x
= 3xy/x - 2y^2/x
= 3y - 2y^2/x

2)
(3x^2y - 4xy^2 + 6x^3y^3)/xy
= xy(3x - 4y + 6x^2y^2)/xy (cancel out xy)
= 3x - 4y + 6x^2y^2
= 6x^2y^2 + 3x - 4y (arrangement)

3)
(7x^3y^2 - 14x^5y^3 + 28x^8y^5 - 21x^7y^6)/7x^3y^2
= 7x^3y^2(1 - 2x^2y + 4x^5y^3 - 3x^4y^4)/7x^3y^3 (cancel out 7x^3y^2)
= 1 - 2x^2y + 4x^5y^3 - 3x^4y^4
= 4x^5y^3 - 3x^4y^4 - 2x^2y + 1

4)
(-15x^2y^3z^3 + 20x^3y^2z^2 - 5x^4y^6z^4 - 10x^2y^2z^3)/5x^2y^2z^2
= 5x^2y^2z^2(-3yz + 4x - x^2y^4z^2 + 2z)/5x^2y^2z^2 (cancel out 5x^2y^2z^2)
= -3yz + 4x - x^2y^4z^2 + 2z
= -x^2y^4z^2 - 3yz + 4x + 2z

5)
(15a^6n - 24a^8n + 6a^5n)/3a^4n
= 3a^5n(5a - 8a^3 + 2)/3a^4n
= (3a^5n/3a^4n)(-8a^3 + 5a + 2)
= a(-8a^3 + 5a + 2)

6)
(6x^4 + 2x^3 - 16x^2 - 4x)/(-2x)
= -2x(-3x^3 - x^2 + 8x + 2)/(-2x) (cancel out -2x)
= -3x^3 - x^2 + 8x + 2

For each of these, split the fraction into as many fractions as number of terms in the numerator
Use laws of indices and some numerical cancellations and simplify that's all.
(9xy^2 - 6y^3) / 3x
= (9xy^2/3x) - (6y^3/3x)
=3y^2 - 2y^2/x, in the first 3x cancells and in the second only 3 cancells.
Now you do the remaining to gain confidence.