$(x^3y^3 + 1) \, dx + x^4y^2 \, dy = 0$

$x^3y^3 \, dx + dx + x^4y^2 \, dy = 0$

$(x^3y^3 \, dx + x^4y^2 \, dy) + dx = 0$

$x^3y^2(y \, dx + x \, dy) + dx = 0$

$x^3y^2 \, d(xy) + dx = 0$

Divide by x both sides

$x^2y^2 \, d(xy) + \dfrac{dx}{x} = 0$

$\displaystyle \int (xy)^2 \, d(xy) + \int \dfrac{dx}{x} = 0$

$\frac{1}{3}(xy)^3 + \ln x = -\ln c$

$x^3y^3 + 3\ln x = -3\ln c$

$x^3y^3 = -3\ln c - 3\ln x$

$x^3y^3 = -3(\ln c + \ln x)$

$x^3y^3 = -3\ln | cx |$ *answer*