Show that the general solution of the differential equation $\frac{{dy}}{{dx}} + \frac{{{y^2} + y + 1}}{{{x^2} + x + 1}} = 0$ is given by $(x + y + 1) = A(1 - x - y - 2xy)$, where $A$ is parameter

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

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

$\Rightarrow$ $\frac{{dy}}{{{y^2} + y + 1}} = \frac{{ - dx}}{{{x^2} + x + 1}}$
$\Rightarrow$ $\frac{{dy}}{{{y^2} + y + 1}} + \frac{{dx}}{{{x^2} + x + 1}} = 0$
Integrating both sides,

we get:
$\int {\frac{{dy}}{{{y^2} + y + 1}}} + \int {\frac{{dx}}{{{x^2} + x + 1}}} = {\rm{C}}$
$\Rightarrow$ $\int {\frac{{dy}}{{{{\left( {y + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} + \int {\frac{{dx}}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} = {\rm{C}}$

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

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

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

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

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

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

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

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

$\Rightarrow$ $x + y + 1 = A(1 - x - y - 2xy)$,

where $A = \frac{{2B}}{{\sqrt 3 }}$

Hence, the given result is proved.