$(2xy - 3x^2) \, dx + (x^2 + y) \, dy = 0$
$M = 2xy - 3x^2$
$N = x^2 + y$
Test for exactness
$\dfrac{\partial M}{\partial y} = 2x$
$\dfrac{\partial N}{\partial x} = 2x$
Exact!
Let
$\dfrac{\partial F}{\partial x} = M$
$\dfrac{\partial F}{\partial x} = 2xy - 3x^2$
$\partial F = (2xy - 3x^2) \, \partial x$
Integrate partially in x, holding y as constant
$\displaystyle \int \partial F = \int (2xy - 3x^2) \, \partial x$
$F = x^2y - x^3 + f(y)$ → Equation (1)
Differentiate partially in y, holding x as constant
$\dfrac{\partial F}{\partial y} = x^2 + f'(y)$
Let
$\dfrac{\partial F}{\partial y} = N$
$x^2 + f'(y) = x^2 + y$
$f'(y) = y$
Integrate partially in y, holding x as constant
$\displaystyle \int f'(y) = \int y \, \partial y$
$f(y) = \frac{1}{2}y^2$
Substitute f(y) to Equation (1)
$F = x^2y - x^3 + \frac{1}{2}y^2$
Equate F to c
$F = c$
$x^2y - x^3 + \frac{1}{2}y^2 = c$ answer
