$3(3x^2 + y^2) \, dx - 2xy \, dy = 0$
Let
$y = vx$
$dy = v \, dx + x \, dv$
Substitute,
$3(3x^2 + v^2x^2) \, dx - 2vx^2 (v \, dx + x \, dv) = 0$
$3(3 + v^2)x^2 \, dx - 2vx^2 (v \, dx + x \, dv) = 0$
Divide by x2,
$3(3 + v^2) \, dx - 2v (v \, dx + x \, dv) = 0$
$9 \, dx + 3v^2 \, dx - 2v^2 \, dx - 2vx \, dv = 0$
$9 \, dx + v^2 \, dx - 2vx \, dv = 0$
$(9 + v^2) \, dx - 2vx \, dv = 0$
$\dfrac{(9 + v^2) \, dx}{x(9 + v^2)} - \dfrac{2vx \, dv}{x(9 + v^2)} = 0$
$\dfrac{dx}{x} - \dfrac{2v \, dv}{9 + v^2} = 0$
$\displaystyle \int \dfrac{dx}{x} - \int \dfrac{2v \, dv}{9 + v^2} = 0$
$\ln x - \ln (9 + v^2) = \ln c$
$\ln \dfrac{x}{9 + v^2} = \ln c$
$\dfrac{x}{9 + v^2} = c$
$x = c(9 + v^2)$
From
$y = vx$
$v = \dfrac{y}{x}$
Thus,
$x = c \left( 9 + \dfrac{y^2}{x^2} \right)$
$x = c \left( \dfrac{9x^2 + y^2}{x^2} \right)$
$x^3 = c(9x^2 + y^2)$ answer