Step 1: Rewrite the given differential equation by separating variables.

Given:

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

Rearranging:

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

Separating variables:

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

Step 2: Integrate both sides.

Note that

$$y^2 + y + 1 = \left(y + \frac{1}{2}\right)^2 + \frac{3}{4} = \left(y + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$$

Similarly,

$$x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$$

Using $\int \frac{dt}{t^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{t}{a}$:

$$\frac{2}{\sqrt{3}}\tan^{-1}\frac{2y + 1}{\sqrt{3}} = -\frac{2}{\sqrt{3}}\tan^{-1}\frac{2x + 1}{\sqrt{3}} + C$$

Step 3: Simplify the equation.

$$\tan^{-1}\frac{2y + 1}{\sqrt{3}} + \tan^{-1}\frac{2x + 1}{\sqrt{3}} = C_1$$

Using $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab}$:

$$\tan^{-1}\left(\frac{\frac{2y+1}{\sqrt{3}} + \frac{2x+1}{\sqrt{3}}}{1 - \frac{(2y+1)(2x+1)}{3}}\right) = C_1$$

Step 4: Take tangent on both sides and simplify.

Taking tangent:

$$\frac{\frac{2x+2y+2}{\sqrt{3}}}{\frac{3-(4xy+2x+2y+1)}{3}} = A$$

(where $A = \tan C_1$)

$$\frac{2(x+y+1)}{\sqrt{3}} \cdot \frac{3}{2-4xy-2x-2y} = A$$ $$\frac{\sqrt{3}(x+y+1)}{1-x-y-2xy} = A$$

Step 5: Rearrange to get the required form.

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

Let $\frac{A}{\sqrt{3}} = A$ (renaming the constant):

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

This is the general solution of the given differential equation, where $A$ is an arbitrary parameter.