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 a parameter.

Given differential equation: $\frac{dy}{dx} + \frac{y^{2}+y+1}{x^{2}+x+1} = 0$
Separating the variables,we get: $\frac{dy}{y^{2}+y+1} = -\frac{dx}{x^{2}+x+1}$
$\Rightarrow \frac{dy}{y^{2}+y+1} + \frac{dx}{x^{2}+x+1} = 0$
Integrating both sides: $\int \frac{dy}{(y+\frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2}} + \int \frac{dx}{(x+\frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2}} = C$
Using the formula $\int \frac{du}{u^{2}+a^{2}} = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + C$,we get:
$\frac{2}{\sqrt{3}} \tan^{-1}(\frac{2y+1}{\sqrt{3}}) + \frac{2}{\sqrt{3}} \tan^{-1}(\frac{2x+1}{\sqrt{3}}) = C$
$\Rightarrow \tan^{-1}(\frac{2y+1}{\sqrt{3}}) + \tan^{-1}(\frac{2x+1}{\sqrt{3}}) = \frac{\sqrt{3}C}{2} = K$ (where $K$ is a constant)
Taking $\tan$ on both sides and using $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:
$\frac{\frac{2y+1}{\sqrt{3}} + \frac{2x+1}{\sqrt{3}}}{1 - (\frac{2y+1}{\sqrt{3}})(\frac{2x+1}{\sqrt{3}})} = \tan K = A'$
$\Rightarrow \frac{2x+2y+2}{\sqrt{3}} = A'(1 - \frac{4xy+2x+2y+1}{3})$
$\Rightarrow \frac{2(x+y+1)}{\sqrt{3}} = A'(\frac{3-4xy-2x-2y-1}{3}) = A'(\frac{2-2x-2y-4xy}{3})$
$\Rightarrow x+y+1 = A(1-x-y-2xy)$,where $A$ is a new constant.