Differential Equation $2y \, dx+x(x^2 \ln y -1) \, dy = 0$

2ydx+x (x²lny-1)dy=0

$2y \, dx+x(x^2 \ln y -1) \, dy = 0$

$dx + x^3 \left( \dfrac{\ln y}{y} \right) \, dy - \dfrac{x}{y} \, dy = 0$

$dx - \dfrac{1}{y} \, x \, dy = -\dfrac{\ln y}{y} \, x^3 \, dy$   ←   Bernoulli's Equation
 

Integrating factor
$\displaystyle u = e^{(1 - n) \int P(y) \, dy} = e^{-2\int (-1/y) \, dy} = e^{2\int (dy/y)}$

$u = e^{2\ln y} = e^{\ln y^2} = y^2$
 

The solution to Bernoulli's Equation is:
$\displaystyle x^{1 - n} u = (1 - n) \int Q(y) \, u \, dy + C$
 

Hence,
$\displaystyle x^{-2} y^2 = -2 \int \left( -\dfrac{\ln y}{y} \right)(y^2) \, dy + C$

$\displaystyle x^{-2} y^2 = 2 \int y \ln y \, dy + C$
 

Thus,
$x^{-2} y^2 = 2 \left( \frac{1}{2}y^2 \ln y - \frac{1}{4}y^2 \right) + C$

$x^{-2} y^2 = y^2 \ln y - \frac{1}{2}y^2 + C$

$y^2 \left( \dfrac{1}{x^2} - \ln y + \dfrac{1}{2} \right) = C$   ←   answer
 

Kindly double-check my solution.