Step 1:

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

This can be written as

$\large\frac{dx}{dy}=\frac{-(x^2-xy+y^2)}{y^2}$

Dividing by $y^2$ we get,

$\large\frac{dx}{dy}=\frac{-x^2}{y^2}+\frac{x}{y}$$-1$

This is clearly a homologous differential equation we can solve by substituting $x=vy$ and $\large\frac{dy}{dx}$$=v+y\large\frac{dv}{dy}$

$v+y \large\frac{dv}{dy}=\frac{-v^2y^2}{y^2}+\frac{vy}{y}-1$

=>$y.\large\frac{dv}{dy}$$=-v^2+v-1-v$

=>$y.\large\frac{dv}{dy}$$=-(v^2+1)$

Now seperating the variables we get,

$\large\frac{dv}{v^2+1}=-\frac{dy}{y}$

Step 2:

Now integrating on both sides we get,

$\int\large\frac{dv}{v^2+1}=-\int\frac{dy}{y}$

$\int\large\frac{dv}{v^2+1}$ is of the form $ \int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1}\bigg(\large\frac{x}{a}\bigg)+c$

Hence $\tan ^{-1}(v)=-\log y+c$

Substituting for $v=x/y$ we get,

$\tan ^{-1}\bigg(\large\frac{x}{y}\bigg)$$=-\log y+c$

Hence the required solution is $\tan ^{-1}\bigg(\large\frac{x}{y}\bigg)$$+\log y=c$