$(y^2 - 2xy + 6x) \, dx - (x^2 - 2xy + 2) \, dy = 0$
$M = y^2 - 2xy + 6x$
$N = -x^2 + 2xy - 2$
Test for exactness
$\dfrac{\partial M}{\partial y} = 2y - 2x$
$\dfrac{\partial N}{\partial x} = -2x + 2y$
Exact!
Let
$\dfrac{\partial F}{\partial x} = M$
$\dfrac{\partial F}{\partial x} = y^2 - 2xy + 6x$
$\partial F = (y^2 - 2xy + 6x) \, \partial x$
Integrate partially in x, holding y as constant
$\displaystyle \int \partial F = \int (y^2 - 2xy + 6x) \, \partial x$
$F = xy^2 - x^2y + 3x^2 + f(y)$ → Equation (1)
Differentiate partially in y, holding x as constant
$\dfrac{\partial F}{\partial y} = 2xy - x^2 + f'(y)$
Let
$\dfrac{\partial F}{\partial y} = N$
$2xy - x^2 + f'(y) = -x^2 + 2xy - 2$
$f'(y) = -2$
Integrate partially in y, holding x as constant
$\displaystyle \int f'(y) = -2 \int \partial y$
$f(y) = -2y$
Substitute f(y) to Equation (1)
$F = xy^2 - x^2y + 3x^2 - 2y$
Equate F to c
$F = c$
$xy^2 - x^2y + 3x^2 - 2y = c$ answer
