$M~dx + N~dy = 0$
$y(2x - y + 1)~dx + x(3x - 4y + 3)~dy = 0$
$M = y(2x - y + 1) = 2xy - y^2 + y$
$N = x(3x - 4y + 3) = 3x^2 - 4xy + 3x$
$\dfrac{\partial M}{\partial y} = 2x - 2y + 1$
$\dfrac{\partial N}{\partial x} = 6x - 4y + 3$
$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = (2x - 2y + 1) - (6x - 4y + 3)$
$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = -4x + 2y - 2$
$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = -2(2x - y + 1)$
$\dfrac{1}{N}\left( \dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} \right) = \dfrac{-2(2x - y + 1)}{x(3x - 4y + 3)}$ → neither a function of x alone nor y alone
$\dfrac{1}{M}\left( \dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} \right) = \dfrac{-2(2x - y + 1)}{y(2x - y + 1)} = \dfrac{-2}{y}$ → a function of y alone
$f(y) = \dfrac{-2}{y}$
Integrating factor
$u = e^{-\int f(y)~dy} = e^{-\int (-2/y)~dy} = e^{2\int (dy/y)}$
$u = e^{2\ln y} = e^{\ln y^2} = y^2$
Thus,
$uM~dx + uN~dy = 0$
$(y^2)y(2x - y + 1)~dx + (y^2)x(3x - 4y + 3)~dy = 0$
$(2xy^3 - y^4 + y^3)~dx + (3x^2y^2 - 4xy^3 + 3xy^2)~dy = 0$
$(3xy^2~dy + y^3~dx) - (4xy^3~dy + y^4~dx) + (3x^2y^2~dy + 2xy^3~dx) = 0$
$[ \, x(3y^2~dy) + y^3~dx \, ] - [ \, x(4y^3~dy) + y^4~dx \, ] + [ \, x^2(3y^2~dy) + y^3(2x~dx) \, ] = 0$
$d(xy^3) - d(xy^4) + d(x^2y^3) = 0$
$\displaystyle \int d(xy^3) - \int d(xy^4) + \int d(x^2y^3) = 0$
$xy^3 - xy^4 + x^2y^3 = c$
$xy^3(1 - y + x) = c$
$xy^3(x - y + 1) = c$ answer
