Answer
$3y^{4}-2y^{3}+\frac{1}{3}y^{2}$
Work Step by Step
$\frac{1}{3}y^{2}(9y^{2}-6y+1)=(\frac{1}{3}y^{2})(9y^{2})+(\frac{1}{3}y^{2})(-6y)+(\frac{1}{3}y^{2})(1)=3y^{2+2}+(-2y^{2+1})+\frac{1}{3}y^{2}=3y^{4}+(-2y^{3})+\frac{1}{3}y^{2}=3y^{4}-2y^{3}+\frac{1}{3}y^{2}$
