First we start out by factoring everything we can so our original problem changes to become : [tex] \frac{y^2}{y-3} [/tex] x [tex] \frac{(y-3)(y+2)}{y(y+1)} [/tex]
Next we cancel terms so since to have a (y-3) on top and the bottom they cancel out leaving you with : [tex] \frac{y^2(y+2)}{y(y+1)} [/tex]
Again with a lone y on top and bottom we cancel terms leaving your answer as : [tex] \frac{y(y+2)}{y+1} [/tex]