differential equations: $y(9x - 2y)dx - x(6x - y)dy = 0$

$y(9x - 2y) \, dx - x(6x - y) \, dy = 0$

$vx(9x - 2vx) \, dx - x(6x - vx)(v \, dx + x \, dv) = 0$

$vx(9x - 2vx) \, dx - vx(6x - vx)\, dx - x^2(6x - vx) \, dv = 0$

$vx(3x - 2vx) \, dx - x^2(6x - vx) \, dv = 0$

$vx^2(3 - 2v) \, dx - x^3(6 - v) \, dv = 0$

$\dfrac{x^2 \, dx}{x^3} - \dfrac{(6 - v) \, dv}{v(3 - 2v)} = 0$

$\dfrac{dx}{x} - \dfrac{(6 - v) \, dv}{v(3 - 2v)} = 0$

$\dfrac{dx}{x} - \left( \dfrac{2}{v} + \dfrac{3}{3 - 2v} \right) \, dv = 0$

$\ln x - 2 \ln v + \frac{3}{2} \ln (3 - 2v) = ln c$

$\ln x - \ln v^2 + \ln (3 - 2v)^{3/2} = ln c$

$\ln \dfrac{x(3 - 2v)^{3/2}}{v^2} = ln c$

$\dfrac{x(3 - 2v)^{3/2}}{v^2} = c$

$\dfrac{x\left( 3 - \dfrac{2y}{x} \right)^{3/2}}{\dfrac{y^2}{x^2}} = c$

$\dfrac{x^3\left( \dfrac{3x - 2y}{x} \right)^{3/2}}{y^2} = c$

$\dfrac{x^3\left[ \dfrac{(3x - 2y)^{3/2}}{x^{3/2}} \right]}{y^2} = c$

$\dfrac{x^{3/2}(3x - 2y)^{3/2}}{y^2} = c$

$x^{3/2}(3x - 2y)^{3/2} = cy^2$           answer

Corrections:
$y(9x - 2y) \, dx - x(6x - y) \, dy = 0$

$vx(9x - 2vx) \, dx - x(6x - vx)(v \, dx + x \, dv) = 0$

$vx(9x - 2vx) \, dx - vx(6x - vx)\, dx - x^2(6x - vx) \, dv = 0$

$vx(3x - vx) \, dx - x^2(6x - vx) \, dv = 0$

$vx^2(3 - v) \, dx - x^3(6 - v) \, dv = 0$

$\dfrac{x^2 \, dx}{x^3} - \dfrac{(6 - v) \, dv}{v(3 - v)} = 0$

$\dfrac{dx}{x} - \dfrac{(6 - v) \, dv}{v(3 - v)} = 0$

$\dfrac{dx}{x} - \left( \dfrac{2}{v} + \dfrac{1}{3 - v} \right) \, dv = 0$

$\ln x - 2 \ln v + \ln (3 - v) = ln c$

$\ln x - \ln v^2 + \ln (3 - v) = ln c$

$\ln \dfrac{x(3 - v)}{v^2} = ln c$

$\dfrac{x(3 - v)}{v^2} = c$

$\dfrac{x\left( 3 - \dfrac{y}{x} \right)}{\dfrac{y^2}{x^2}} = c$

$\dfrac{x^3\left( \dfrac{3x - y}{x} \right)}{y^2} = c$

$\dfrac{x^2(3x - y)}{y^2} = c$

$x^2(3x - y) = cy^2$           answer