Answer
$$-\frac{\sqrt{5-y^{2}}}{5 y}+C$$
Work Step by Step
Given $$ \int \frac{d y}{y^{2} \sqrt{5-y^{2}}} $$
Let $$ y= \sqrt{5} \sin \theta ,\ \ \ dy= \sqrt{5}\cos \theta d \theta $$
Then
\begin{aligned}
\int \frac{d y}{y^{2} \sqrt{5-y^{2}}} &=\int \frac{\sqrt{5} \cos \theta d \theta}{(\sqrt{5} \sin \theta)^{2} \sqrt{5-(\sqrt{5} \sin \theta)^{2}}} \\
&=\int \frac{\sqrt{5} \cos \theta d \theta}{5 \sin ^{2} \theta \sqrt{5-5 \sin ^{2} \theta}} \\
&=\int \frac{\sqrt{5} \cos \theta d \theta}{5 \sqrt{5} \sin ^{2} \theta \sqrt{1-\sin ^{2} \theta}} \\
&=\frac{1}{5} \int \frac{\cos \theta d \theta}{\sin ^{2} \theta \sqrt{\cos ^{2} \theta}} \\
&=\frac{1}{5} \int \frac{\cos \theta d \theta}{\sin ^{2} \theta \cos \theta}\\
&=\frac{1}{5} \int \csc ^{2} \theta d \theta\\
&=\frac{1}{5}(-\cot \theta)+C\\
&= -\frac{\sqrt{5-y^{2}}}{5 y}+C
\end{aligned}
