Answer
$$\frac{1}{\left(x+y\right)^3\left(3x+y\right)}$$
Work Step by Step
Recall, one never divides by a fraction. Rather, if we want to divide by a fraction, we multiply by the reciprocal of the fraction. Recall, the reciprocal of a fraction is that fraction with the numerator and the denominator switched. Thus, we find:
$$ \frac{3x^2-2xy-y^2}{\left(x^2-y^2\right)\left(3x^2+4xy+y^2\right)^2}\\ \frac{\left(3x+y\right)\left(x-y\right)}{\left(x+y\right)^2\left(3x+y\right)^2\left(x^2-y^2\right)}\\ \frac{1}{\left(x+y\right)^3\left(3x+y\right)}$$
