Answer
$\displaystyle \frac{3}{y}$
Work Step by Step
Dividing with $\displaystyle \frac{P}{Q}$ equals multiplying with the reciprocal, $\displaystyle \frac{Q}{P}.$
$\displaystyle \frac{3y^{2}-12}{y^{2}+4y+4}\div\frac{y^{3}-2y^{2}}{y^{2}+2y}=\frac{3y^{2}-12}{y^{2}+4y+4}\cdot\frac{y^{2}+2y}{y^{3}-2y^{2}}\qquad$
... factor what you can
$3y^{2}-12=3(y^{2}-4)=3(y+2)(y-2)$
$y^{2}+4y+4=(y+2)^{2}$
$y^{2}+2y=y(y+2)$
$y^{3}-2y^{2}=y^{2}(y-2)$
$=\displaystyle \frac{3(y+2)(y-2)}{(y+2)^{2}}\cdot\frac{y(y+2)}{y^{2}(y-2)} \qquad$... divide out the common factors
$=\displaystyle \frac{3(1)(1)}{(1)^{1}}\cdot\frac{1(1)}{y^{1}(1)} $
= $\displaystyle \frac{3}{y}$
