Factorize the denominator.

The denominator is $9x^2 - y^2$. This is a difference of squares, which follows the formula $a^2 - b^2 = (a-b)(a+b)$.
Here, $a = 3x$ and $b = y$.

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

The numerator is $3x^2 - 4xy - y^2$. We look for two binomials $(Ax+By)(Cx+Dy)$ such that their product is $3x^2 - 4xy - y^2$.
Expanding $(Ax+By)(Cx+Dy) = ACx^2 + (AD+BC)xy + BDy^2$.
We need to find integer coefficients $A, B, C, D$ such that:
$AC = 3$
$BD = -1$
$AD+BC = -4$

Possible integer pairs for $(A,C)$: $(1,3)$ or $(3,1)$.
Possible integer pairs for $(B,D)$: $(1,-1)$ or $(-1,1)$.

Let's test all combinations:

• If $(A,C) = (1,3)$ and $(B,D) = (1,-1)$: $(x+y)(3x-y) = 3x^2 - xy + 3xy - y^2 = 3x^2 + 2xy - y^2$. (Incorrect middle term)
• If $(A,C) = (1,3)$ and $(B,D) = (-1,1)$: $(x-y)(3x+y) = 3x^2 + xy - 3xy - y^2 = 3x^2 - 2xy - y^2$. (Incorrect middle term)
• If $(A,C) = (3,1)$ and $(B,D) = (1,-1)$: $(3x+y)(x-y) = 3x^2 - 3xy + xy - y^2 = 3x^2 - 2xy - y^2$. (Incorrect middle term)
• If $(A,C) = (3,1)$ and $(B,D) = (-1,1)$: $(3x-y)(x+y) = 3x^2 + 3xy - xy - y^2 = 3x^2 + 2xy - y^2$. (Incorrect middle term)

Since none of these combinations result in the middle term $-4xy$, the numerator $3x^2 - 4xy - y^2$ is not factorable over integers.

Since the numerator cannot be factored over integers, the expression is presented with the denominator factored.

The fully factorized expression is:

$$\frac{3x^2 - 4xy - y^2}{(3x - y)(3x + y)}$$

The numerator $3x^2 - 4xy - y^2$ is not factorable over integers.