$y(x^4 - y^2) \, dx + x(x^4 + y^2) \, dy = 0$
$x^4y \, dx - y^3 \, dx + x^5 \, dy + xy^2 \, dy = 0$
$(x^4y \, dx + x^5 \, dy) + (xy^2 \, dy - y^3 \, dx) = 0$
$x^4(y \, dx + x \, dy) + y^2(x \, dy - y \, dx) = 0$
$(y \, dx + x \, dy) + \dfrac{y^2(x \, dy - y \, dx)}{x^4} = 0$
$(y \, dx + x \, dy) + \dfrac{y^2}{x^2} \left( \dfrac{x \, dy - y \, dx}{x^2} \right) = 0$
$(y \, dx + x \, dy) + \left( \dfrac{y}{x} \right)^2 \left( \dfrac{x \, dy - y \, dx}{x^2} \right) = 0$
$d(xy) + \left( \dfrac{y}{x} \right)^2 d\left( \dfrac{y}{x} \right) = 0$
$\displaystyle \int d(xy) + \int \left( \dfrac{y}{x} \right)^2 d\left( \dfrac{y}{x} \right) = 0$
$xy + \dfrac{1}{3}\left( \dfrac{y}{x} \right)^3 = \dfrac{c}{3}$
$xy + \dfrac{y^3}{3x^3} = \dfrac{c}{3}$
$3x^4y + y^3 = cx^3$
$y(3x^4 + y^2) = cx^3$ answer
