[tex]y'=\dfrac{-3\,{y}^{2}+x\,y+{x}^{2}}{{x}^{2}-4\,x\,y}\Leftrightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{-3\,{y}^{2}+x\,y+{x}^{2}}{{x}^{2}-4\,x\,y}[/tex]

Умножим всё на [tex]\mathrm{d}x[/tex] и сделаем замену [tex]u=\frac{y}{x}[/tex], тогда [tex]y=ux[/tex], а [tex]\mathrm{d}y=u\mathrm{d}x+x\mathrm{d}u[/tex], получаем

[tex]u\,\mathrm{d}x+x\,\mathrm{d}u=\dfrac{\left(-3\,{u}^{2}+u+1\right)\,\mathrm{d}x}{1-4\,u}\Leftrightarrow x\,\mathrm{d}u=\left(\dfrac{-3\,{u}^{2}+u+1}{1-4\,u}-u\right)\,\mathrm{d}x[/tex]

[tex]-\dfrac{\left(4\,u-1\right)\,\mathrm{d}u}{{u}^{2}+1}=\dfrac{\mathrm{d}x}{x}\Rightarrow \int{-\dfrac{4\,u-1}{{u}^{2}+1}}{\;\mathrm{d}u}=\int{\dfrac{1}{x}}{\;\mathrm{d}x}\Rightarrow \\\Rightarrow \mathrm{arctg}\left(u\right)-2\,\ln\left({u}^{2}+1\right)=\ln\left(x\right)+C\Rightarrow \\\Rightarrow\mathrm{arctg}\left(\frac{y}{x}\right)-2\,\ln\left({y}^{2}+{x}^{2}\right)=C-3\,\ln\left(x\right)[/tex]