problem:
6y²dx - x(2x³+y)dy = 0
n = ?
P = ?
Q = ?
V = ?
general solution = ?
You first need to re-write this equation in Bernoulli form
$\displaystyle \frac{dy}{dx} + P(x)y = Q(x) y^n$. Isolating $\displaystyle \frac{dy}{dx} $ gives
$\displaystyle \frac{dy}{dx} = \frac{6y^2}{x(2x^3+y)} $
which isn't of the right form but isolating $\displaystyle \frac{dx}{dy} $
gives
$\displaystyle \frac{dx}{dy} - \frac{x}{6y} = \frac{x^4}{3y^2} $ where $\displaystyle P = -\frac{1}{6y}, \;\;\ Q = \frac{1}{3y^2} \,\,\, n = 4 $. I don't know what your V is?
Before we discuss the solution, do you know how to solve Bernoulli equations?
Before you find your integrating factor, you will need to linearize the equation
so
$\displaystyle \frac{dx}{dy} - \frac{x}{6y} = \frac{x^4}{3y^2}$
becomes
$\displaystyle \frac{1}{x^4} \frac{dx}{dy} - \frac{1}{x^3} \frac{1}{6y} = \frac{1}{3y^2} $
and then let $\displaystyle u = \frac{1}{x^3}$. See how you go from here.