Answer
$$\ln \left|\frac{y}{3}+\frac{\sqrt{y^{2}-9}}{3}\right|+C$$
Work Step by Step
Given $$\int \frac{d y}{\sqrt{y^{2}-9}}$$
Let
$$y=3\sec u\ \ \ \ \ \ \to \ \ \ \ \ dy=3\sec u\tan udu $$
Then
\begin{align*}
\int \frac{d y}{\sqrt{y^{2}-9}}&=\int \frac{3\sec u\tan udu}{\sqrt{9\sec^2u-9}}\\
&=\int \frac{3\sec u\tan udu}{\sqrt{9\tan^2u}}\\
&=\int\sec udu\\
&=\ln|\sec u+\tan u|+C\\
&=\ln \left|\frac{y}{3}+\frac{\sqrt{y^{2}-9}}{3}\right|+C
\end{align*}
