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

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 x^{2},

$3(3 + v^2) \, dx - 2v (v \, dx + x \, dv) = 0$

$9 \, dx + 3v^2 \, dx - 2v^2 \, dx - 2vx \, dv = 0$

$9 \, dx + v^2 \, dx - 2vx \, dv = 0$

$(9 + v^2) \, dx - 2vx \, dv = 0$

$\dfrac{(9 + v^2) \, dx}{x(9 + v^2)} - \dfrac{2vx \, dv}{x(9 + v^2)} = 0$

$\dfrac{dx}{x} - \dfrac{2v \, dv}{9 + v^2} = 0$

$\displaystyle \int \dfrac{dx}{x} - \int \dfrac{2v \, dv}{9 + v^2} = 0$

$\ln x - \ln (9 + v^2) = \ln c$

$\ln \dfrac{x}{9 + v^2} = \ln c$

$\dfrac{x}{9 + v^2} = c$

$x = c(9 + v^2)$

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*