dy/dx = (x^2)+(8y^2)/(3xy)? I know that you have to name a parmater v that takes the value y/x and substitute it back in but I'm lost after that.
If your differential equation is
$\displaystyle \frac{dy}{dx} = \frac{x^2 + 8y^2}{3xy}$
You can do a number of things. First the equation is homogeneous so if you let
$\displaystyle v = \frac{y}{x}$ as you have said, then $\displaystyle y = x v$ then $\displaystyle y' = x v' + v$, substitute giving
$\displaystyle x v' + v = \frac{x^2 + 8 x^2 v^2}{3x^2 v}$
can x's from the right and solve for v' giving something that is separable. The equation is also Bernoulli.
Ahh. I see.
Your equation can be written:
$\displaystyle \frac{dy}{dx} = \frac{x^2}{3xy}+\frac{8y^2}{3xy} $
Which gives:
$\displaystyle \frac{dy}{dx} = \frac{x}{3y}+\frac{8y}{3x} $
Giving:
$\displaystyle \frac{dy}{dx} = \frac{1}{3}(\frac{y}{x})^{-1}+\frac{8}{3}\frac{y}{x} $
$\displaystyle \text{Let} v = \frac{y}{x} $
$\displaystyle y = vx $
Hence
$\displaystyle \frac{dy}{dx} = v +x\frac{dv}{dx} $
Hence:
$\displaystyle v +x\frac{dv}{dx} = \frac{1}{3}(v)^{-1}+\frac{8}{3}v $