The general solution of $y^2dx+(x^2-xy+y^2)dy = 0 $, is

$log y = tan^{-1} \frac{y}{x}+C$

The correct answer is option (4) : $log y = tan^{-1} \frac{y}{x}+C$

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

$⇒\frac{dy}{dx} = - \frac{y^2}{x^2-xy+y^2}$

Putting $ y = vx $ and $\frac{dy}{dx} = v + x\frac{dv}{dx} $, we get

$v+x\frac{dv}{dx}= -\frac{v^2}{1-v+v^2}$

$⇒x\frac{dv}{dx}=\frac{-v-v^3}{v^2-v+1}$

$⇒\frac{v^2-v+1}{v(v^2+1)}=-\frac{dx}{x}$

$⇒\left(\frac{1}{v}-\frac{1}{v^2+1}\right)dv=-\frac{dx}{x}$

On integrating, we get

$log v - tan^{-1} v = - log  x + C$

$⇒log (\frac{y}{x} ) - tan^{-1} \frac{y}{x} = - log x + C$

$⇒log y = tan^{-1} \frac{y}{x} + C$