Solving $ \frac{dy}{dx} = y^2 - 9$ $ \frac{dy}{dx} = y^2 - 9$
This is separable so I rewrite it as $ \frac{1}{(y^2 - 9)}dy = dx$ then I get
$$\int \frac{1}{(y^2 - 9)}dy = \int dx = \int 1dx = x + c,$$ for some $c \in \mathbb{R} $
The left hand side:
$$\int \frac{1}{(y^2 - 9)}dy = \int \frac{1}{(y+3)(y-3)}dy = \int \left( \frac{A}{y+3} + \frac{B}{ y-3}\right) \,dy $$
We have $A(y-3) + B(y+3) = 1 \implies A =-\frac{1}{6}, B = \frac{1}{6}$
So we get
$$ \int \left(\frac{A}{y+3} + \frac{B}{ y-3}\right) \, dy = \int \left (\frac{-\frac{1}{6}}{(y+3)} + \frac{\frac{1}{6}}{ (y-3)}\right)dy = \frac{1}{6}\int \left (\frac{1}{ (y-3)} - \frac{1}{ (y+3)}\right) dy = \frac{1}{6} \left(\int \frac{1}{ (y-3)}dy - \int \frac{1}{ (y+3)} dy\right ) = \frac{1}{6}\left ( \ln \left \vert y-3 \right\vert - \ln \left\vert y+3 \right\vert \right ) = \frac{1}{6}\ln \left \vert\frac{y-3}{y+3} \right \vert $$
So I have $\ln \left\vert\frac{y-3}{y+3}\right\vert  = 6x + 6c \implies e^{\ln\left\vert\frac{y-3}{y+3}\right\vert} = \left\vert\frac{y-3}{y+3}\right\vert =e^{6x + 6c} $ 
Is this correct to far? If yes how do I solve this for $y$?

$\frac{y-3}{y+3}=\pm e^{6x}e^{6c}=Ce^{6x}$ so $y-3=(y+3)Ce^{6x}$ Group the $y$ terms together to get $y(1-Ce^{6x})=3+3Ce^{6x}$.  Now divide by $1-Ce^{6x}$.
