Answer
$$ - \frac{{\sqrt {9 - {y^2}} }}{{9y}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dy}}{{{y^2}\sqrt {9 - {y^2}} }}} \cr
& {\text{substitute }}y = 3\sin \theta ,{\text{ }}dy = 3\cos \theta d\theta \cr
& \int {\frac{{dy}}{{{y^2}\sqrt {9 - {y^2}} }}} = \int {\frac{{3\cos \theta d\theta }}{{{{\left( {3\sin \theta } \right)}^2}\sqrt {9 - {{\left( {3\sin \theta } \right)}^2}} }}} \cr
& = \int {\frac{{3\cos \theta d\theta }}{{{{\left( {3\sin \theta } \right)}^2}\sqrt {9 - {{\left( {3\sin \theta } \right)}^2}} }}} \cr
& {\text{simplify}} \cr
& = \int {\frac{{3\cos \theta d\theta }}{{9{{\sin }^2}\theta \sqrt {9\left( {1 - {{\sin }^2}\theta } \right)} }}} \cr
& = \int {\frac{{3\cos \theta d\theta }}{{27{{\sin }^2}\theta \sqrt {{{\cos }^2}\theta } }}} \cr
& = \frac{1}{9}\int {\frac{1}{{{{\sin }^2}\theta }}} d\theta \cr
& = \frac{1}{9}\int {{{\csc }^2}\theta } d\theta \cr
& {\text{find the antiderivative}} \cr
& = - \frac{1}{9}\cot \theta + C \cr
& {\text{write in terms of }}y \cr
& = - \frac{1}{9}\left( {\frac{{\sqrt {9 - {y^2}} }}{y}} \right) + C \cr
& = - \frac{{\sqrt {9 - {y^2}} }}{{9y}} + C \cr} $$
