Let $$y = y(x)$$ be the solution of the differential equation $$\cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx$$, $$0 \leq x \leq \frac{\pi}{2}$$, $$y(0) = 0$$. Then, $$y\left(\frac{\pi}{3}\right)$$ is equal to:

The differential equation is $$\cos x(3\sin x + \cos x + 3)\,dy = (1 + y\sin x(3\sin x + \cos x + 3))\,dx$$.

Let us denote $$u = 3\sin x + \cos x + 3$$ for convenience. Then the equation becomes: $$u\cos x\,dy = (1 + yu\sin x)\,dx$$.

Rearranging: $$u\cos x\,\frac{dy}{dx} - yu\sin x = 1$$, so $$u\left(\cos x\,\frac{dy}{dx} - y\sin x\right) = 1$$.

Notice that $$\frac{d}{dx}(y\cos x) = \cos x\,\frac{dy}{dx} - y\sin x$$. So the equation becomes: $$u \cdot \frac{d}{dx}(y\cos x) = 1$$.

Let $$v = y\cos x$$. Then $$\frac{dv}{dx} = \frac{1}{u} = \frac{1}{3\sin x + \cos x + 3}$$.

So $$v = \int \frac{dx}{3\sin x + \cos x + 3}$$.

Using the Weierstrass substitution $$t = \tan\frac{x}{2}$$, so $$\sin x = \frac{2t}{1+t^2}$$, $$\cos x = \frac{1-t^2}{1+t^2}$$, $$dx = \frac{2\,dt}{1+t^2}$$.

The denominator becomes: $$3 \cdot \frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} + 3 = \frac{6t + 1 - t^2 + 3(1+t^2)}{1+t^2} = \frac{6t + 1 - t^2 + 3 + 3t^2}{1+t^2} = \frac{2t^2 + 6t + 4}{1+t^2}$$.

So the integral becomes: $$\int \frac{1+t^2}{2t^2 + 6t + 4} \cdot \frac{2\,dt}{1+t^2} = \int \frac{2\,dt}{2t^2 + 6t + 4} = \int \frac{dt}{t^2 + 3t + 2} = \int \frac{dt}{(t+1)(t+2)}$$.

By partial fractions: $$\frac{1}{(t+1)(t+2)} = \frac{1}{t+1} - \frac{1}{t+2}$$.

Integrating: $$\ln|t+1| - \ln|t+2| + C = \ln\left|\frac{t+1}{t+2}\right| + C$$.

So $$y\cos x = \ln\left|\frac{\tan\frac{x}{2}+1}{\tan\frac{x}{2}+2}\right| + C$$.

Applying the initial condition $$y(0) = 0$$: at $$x = 0$$, $$\tan 0 = 0$$, $$\cos 0 = 1$$, so $$0 = \ln\left|\frac{1}{2}\right| + C$$, giving $$C = \ln 2$$.

Therefore $$y\cos x = \ln\left|\frac{\tan\frac{x}{2}+1}{\tan\frac{x}{2}+2}\right| + \ln 2 = \ln\left|\frac{2(\tan\frac{x}{2}+1)}{\tan\frac{x}{2}+2}\right|$$.

At $$x = \frac{\pi}{3}$$: $$\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$$ and $$\cos\frac{\pi}{3} = \frac{1}{2}$$.

$$y \cdot \frac{1}{2} = \ln\left(\frac{2\left(\frac{1}{\sqrt{3}}+1\right)}{\frac{1}{\sqrt{3}}+2}\right) = \ln\left(\frac{\frac{2(\sqrt{3}+1)}{\sqrt{3}}}{\frac{1+2\sqrt{3}}{\sqrt{3}}}\right) = \ln\left(\frac{2(\sqrt{3}+1)}{1+2\sqrt{3}}\right)$$.

Rationalizing: $$\frac{2(\sqrt{3}+1)}{2\sqrt{3}+1} \cdot \frac{2\sqrt{3}-1}{2\sqrt{3}-1} = \frac{2(\sqrt{3}+1)(2\sqrt{3}-1)}{12-1} = \frac{2(2\cdot 3 - \sqrt{3} + 2\sqrt{3} - 1)}{11} = \frac{2(5 + \sqrt{3})}{11} = \frac{10 + 2\sqrt{3}}{11}$$.

So $$y = 2\ln\left(\frac{2\sqrt{3}+10}{11}\right) = 2\log_e\left(\frac{2\sqrt{3}+10}{11}\right)$$.

The answer is $$2\log_e\left(\frac{2\sqrt{3}+10}{11}\right)$$, which is Option B.