$xy \, dx - (x^2 + 3y^2) \, dy = 0$
Let
$x = vy$
$dx = v \, dy + y \, dv$
$vy^2(v \, dy + y \, dv) - (v^2y^2 + 3y^2) \, dy = 0$
$vy^2(v \, dy + y \, dv) - y^2(v^2 + 3) \, dy = 0$
$v(v \, dy + y \, dv) - (v^2 + 3) \, dy = 0$
$v^2 \, dy + vy \, dv - v^2 \, dy - 3 \, dy = 0$
$vy \, dv - 3 \, dy = 0$
$v \, dv - \dfrac{3 \, dy}{y} = 0$
$\displaystyle \int v \, dv - 3 \int \dfrac{dy}{y} = 0$
$\frac{1}{2}v^2 - 3 \ln y + 3 \ln c = 0$
$v^2 - 6 \ln y + 6 \ln c = 0$
$v^2 - 6(\ln y - \ln c) = 0$
$v^2 - 6\ln \dfrac{y}{c} = 0$
$v^2 = 6\ln \dfrac{y}{c}$
From
$x = vy$
$v = \dfrac{x}{y}$
Thus,
$\dfrac{x^2}{y^2} = 6 \ln \dfrac{y}{c}$
$x^2 = 6y^2 \ln | \, y/c \, |$ answer
