Integrate using trigonometric substitution. Am I on the right path? I have been trying to solve:
$$\int \frac{\sqrt{x^2-9}}{x^3} dx$$
I am letting $ x = 3\sec \theta$ and so $dx = 3 \sec \theta \tan \theta$ 
So then I have:
$$\int \frac{\sqrt{9\sec^2 \theta - 9}}{27 \sec^3 \theta} dx$$
$$\int \frac{\sqrt{9(\sec^2 \theta - 1)}}{27 \sec^3 \theta} 3 \sec \theta \tan \theta\, d \theta$$
$$ \int \frac{3\tan \theta}{27 \sec^3 \theta} 3 \sec \theta \tan \theta\, d \theta$$
$$ = \int \frac{9 \tan ^2 \theta}{27 \sec ^2 \theta} d \theta$$
$$ = \int \frac{\tan ^2 \theta}{3 \sec ^2 \theta} d \theta$$
$$ \int \frac{\sin^2 \theta}{\cos^2 \theta} \cdot \frac{\cos^2 \theta}{3} d \theta$$
$$ \int \frac{\sin^2 \theta}{3} d \theta$$
$$\frac{1}{3} \int \sin^2 \theta d \theta$$
Am I on the right track?

Yes, your solution so far is correct. Now, to integrate $\int \sin^2{\theta}\,d\theta$, use the half-angle formula for the sine function:
$$
\sin^2{\theta}=\frac{1-\cos(2\theta)}{2}.
$$
Also, a bit later, you're going to need this formula:
$$
\sin{(2\theta)}=2\sin{\theta}\cos{\theta}.
$$
