If $$y = y(x)$$ is the solution of the differential equation $$(1+e^{2x})\frac{dy}{dx} + 2(1+y^2)e^x = 0$$ and $$y(0) = 0$$, then $$6\left(y'(0) + \left(y\left(\log_e \sqrt{3}\right)\right)^2\right)$$ is equal to:

The given differential equation is: $$(1 + e^{2x})\frac{dy}{dx} + 2(1 + y^2)e^x = 0$$. Separating variables gives $$\frac{dy}{1 + y^2} = \frac{-2e^x\,dx}{1 + e^{2x}}$$. Integrating both sides, the left side yields $$\int \frac{dy}{1 + y^2} = \arctan(y)$$, and for the right side, with $$u = e^x$$ so that $$du = e^x\,dx$$, we get $$\int \frac{-2\,du}{1 + u^2} = -2\arctan(u) = -2\arctan(e^x)$$. Therefore, $$\arctan(y) = -2\arctan(e^x) + C$$.

Applying the initial condition $$y(0) = 0$$ yields $$\arctan(0) = -2\arctan(e^0) + C \implies 0 = -2 \times \frac{\pi}{4} + C \implies C = \frac{\pi}{2}$$. Evaluating the original equation at $$x = 0$$, where $$y = 0$$, gives $$(1 + 1)y'(0) + 2(1 + 0)(1) = 0 \implies 2y'(0) + 2 = 0 \implies y'(0) = -1$$.

At $$x = \log_e\sqrt{3}$$, we have $$e^x = \sqrt{3}$$ and thus $$\arctan(e^x) = \arctan(\sqrt{3}) = \dfrac{\pi}{3}$$. It follows that $$\arctan(y) = -2 \times \frac{\pi}{3} + \frac{\pi}{2} = -\frac{2\pi}{3} + \frac{\pi}{2} = -\frac{\pi}{6}$$, so $$y = \tan\left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}$$.

Finally, $$6\left(y'(0) + \left(y\left(\log_e\sqrt{3}\right)\right)^2\right) = 6\left(-1 + \frac{1}{3}\right) = 6 \times \left(-\frac{2}{3}\right) = -4$$. The correct answer is Option C: $$-4$$.