$y^2 d x+\left(x^2-x y-y^2\right) d y=0$
$\Rightarrow \frac{d y}{d x}=\frac{-y^2}{x^2-x y-y^2}=\frac{-\left(\frac{y}{x}\right)^2}{1-\left(\frac{y}{x}\right)-\left(\frac{y}{x}\right)^2}$
Let $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\begin{aligned} & \Rightarrow \quad v+x \frac{d v}{d x}=\frac{-v^2}{1-v-v^2} \\ & \Rightarrow \quad x \frac{d v}{d x}=\frac{-v^2}{1-v-v^2}-v \\ & \Rightarrow \quad x \frac{d v}{d x}=\frac{-v^2-v\left(1-v-v^2\right)}{1-v-v^2} \\ & \Rightarrow \quad \frac{d x}{x}=\frac{v^2+v-1}{v^2-v\left(v^2+v-1\right)} d v=\frac{v^2+v-1}{-v\left(v^2-1\right)} d v\end{aligned}$
$\int \frac{d x}{x}=\int \frac{d v}{-v}-\int \frac{d v}{v^2-1}$
$\ln x=-\ln v-\frac{1}{2} \log \left|\frac{v-1}{v+1}\right|+C$
$\ln x=\ln \left(\frac{x}{y}\right)-\frac{1}{2} \log \left|\frac{y-x}{y+x}\right|+C$
$x=\frac{x}{y} \times \sqrt{\frac{y+x}{x-y}}+C$
at $x=2, y=1 \Rightarrow C=2-2 \sqrt{3}$
$\begin{aligned} & \left.\Rightarrow \quad(x-C)=\frac{x}{\left(\frac{x}{x} y\right.}\right) \\ & \therefore \quad(x+y)=\frac{y^2(x-C)^2(x-y)}{x^2} \\ & \quad=\left(\frac{x-C}{x}\right)^2\left(x y^2-y^3\right)\end{aligned}$
at $(2,1), K=\left(\frac{x-C}{x}\right)^2=\left(\frac{2-2+2 \sqrt{3}}{2}\right)^2=3$.