is $\int_{-1}^{1} \frac{2x}{x^2 - 9}dx = 0$? am I doing this the right way?
$\int_{-1}^{1} \frac{2x}{x^2 - 9}dx = \int_{-1}^{1} \frac{d(x^2 - 9)}{x^2 - 9}dx = \left[\ln|x^2 - 9|\right]_{-1}^{1} = 0$
The calculation is correct. 

You might note instead that in the interval you are integrating over, our function is well-behaved (integrable). It is an odd function, and we are integrating over an interval symmetric about $x=0$. So we can conclude that the integral is $0$ without doing any calculation at all.

Recall that a function $f(x)$ is odd if $f(-x)=-f(x)$ for all $x$. 

You might want to sketch the curve, or have software do it for you. The negative "area" from $-1$ to $0$ exactly cancels the positive area from $0$ to $1$.
Not quite. Use u-substitution and set u=x^2-9.