Answer
$$ \frac{4y^2+6y+9}{8y^2+10y-3}$$
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:
$$ \frac{\left(8y^3-27\right)}{64y^3-1}\times \frac{16y^2+4y+1}{\left(4y^2-9\right)}\\ \frac{\left(2y-3\right)\left(4y^2+6y+9\right)\left(16y^2+4y+1\right)}{\left(4y-1\right)\left(16y^2+4y+1\right)\left(2y+3\right)\left(2y-3\right)}\\ \frac{4y^2+6y+9}{\left(4y-1\right)\left(2y+3\right)}\\ \frac{4y^2+6y+9}{8y^2+10y-3}$$