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