Show that the general solution of the differential equation

$\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0$

$\Rightarrow \frac{d y}{d x}=-\frac{\left(y^{2}+y+1\right)}{x^{2}+x+1}$

$\Rightarrow \frac{d y}{y^{2}+y+1}=\frac{-d x}{x^{2}+x+1}$

$\Rightarrow \frac{d y}{y^{2}+y+1}+\frac{d x}{x^{2}+x+1}=0$

Integrating both sides, we get:

$\int \frac{d y}{y^{2}+y+1}+\int \frac{d x}{x^{2}+x+1}=\mathrm{C}$

$\Rightarrow \int \frac{d y}{\left(y+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}+\int \frac{d x}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=\mathrm{C}$

$\Rightarrow \frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right]+\frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right]=\mathrm{C}$

$\Rightarrow \tan ^{-1}\left[\frac{2 y+1}{\sqrt{3}}\right]+\tan ^{-1}\left[\frac{2 x+1}{\sqrt{3}}\right]=\frac{\sqrt{3} \mathrm{C}}{2}$

$\Rightarrow \tan ^{-1}\left[\frac{\frac{2 y+1}{\sqrt{3}}+\frac{2 x+1}{\sqrt{3}}}{1-\frac{(2 y+1)}{\sqrt{3}} \cdot \frac{(2 x+1)}{\sqrt{3}}}\right]=\frac{\sqrt{3} C}{2}$

$\Rightarrow \tan ^{-1}\left[\frac{\frac{2 x+2 y+2}{\sqrt{3}}}{1-\left(\frac{4 x y+2 x+2 y+1}{3}\right)}\right]=\frac{\sqrt{3} \mathrm{C}}{2}$

$\Rightarrow \tan ^{-1}\left[\frac{2 \sqrt{3}(x+y+1)}{3-4 x y-2 x-2 y-1}\right]=\frac{\sqrt{3} \mathrm{C}}{2}$

$\Rightarrow \tan ^{-1}\left[\frac{\sqrt{3}(x+y+1)}{2(1-x-y-2 x y)}\right]=\frac{\sqrt{3} \mathrm{C}}{2}$

$\Rightarrow \frac{\sqrt{3}(x+y+1)}{2(1-x-y-2 x y)}=\tan \left(\frac{\sqrt{3} C}{2}\right)=B$, where $B=\tan \left(\frac{\sqrt{3} C}{2}\right)$

$\Rightarrow x+y+1=\frac{2 B}{\sqrt{3}}(1-x y-2 x y)$

$\Rightarrow x+y+1=A(1-x-y-2 x y)$, where $A=\frac{2 B}{\sqrt{3}}$

Hence, the given result is proved.