Let the solution curve $$x = x(y)$$, $$0 < y < \frac{\pi}{2}$$, of the differential equation $$(\log_e(\cos y))^2 \cos y \, dx - (1 + 3x \log_e(\cos y)) \sin y \, dy = 0$$ satisfy $$x\left(\frac{\pi}{3}\right) = \frac{1}{2\log_e 2}$$. If $$x\left(\frac{\pi}{6}\right) = \frac{1}{\log_e m - \log_e n}$$, where $$m$$ and $$n$$ are coprime, then $$mn$$ is equal to _____.

We are given the differential equation:

$$(\log_e(\cos y))^2 \cos y \, dx - (1 + 3x \log_e(\cos y)) \sin y \, dy = 0$$

with $$x = x(y)$$, $$0 < y < \dfrac{\pi}{2}$$, and the initial condition $$x\!\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2\log_e 2}$$.

First, we rewrite as $$\dfrac{dx}{dy}$$.

$$(\ln(\cos y))^2 \cos y \, dx = (1 + 3x \ln(\cos y)) \sin y \, dy$$

$$\dfrac{dx}{dy} = \dfrac{(1 + 3x \ln(\cos y)) \tan y}{(\ln(\cos y))^2}$$

Next, we substitute $$t = \ln(\cos y)$$.

Let $$t = \ln(\cos y)$$. Then $$\dfrac{dt}{dy} = -\tan y$$, so $$\tan y \, dy = -dt$$.

By the chain rule:

$$\dfrac{dx}{dt} = \dfrac{dx/dy}{dt/dy} = \dfrac{(1 + 3xt)\tan y / t^2}{-\tan y} = -\dfrac{1 + 3xt}{t^2} = -\dfrac{1}{t^2} - \dfrac{3x}{t}$$

From this, we solve the linear ODE.

$$\dfrac{dx}{dt} + \dfrac{3}{t}\, x = -\dfrac{1}{t^2}$$

Integrating factor: $$e^{\int \frac{3}{t}\, dt} = e^{3\ln|t|} = t^3$$

Multiplying both sides by $$t^3$$:

$$\dfrac{d}{dt}(t^3 x) = -t$$

Integrating:

$$t^3 x = -\dfrac{t^2}{2} + C$$

$$x = -\dfrac{1}{2t} + \dfrac{C}{t^3}$$

Now, we apply the initial condition.

At $$y = \dfrac{\pi}{3}$$: $$\cos\!\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}$$, so $$t = \ln\!\left(\dfrac{1}{2}\right) = -\ln 2$$.

Given $$x\!\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2\ln 2}$$:

$$\dfrac{1}{2\ln 2} = -\dfrac{1}{2(-\ln 2)} + \dfrac{C}{(-\ln 2)^3} = \dfrac{1}{2\ln 2} - \dfrac{C}{(\ln 2)^3}$$

$$0 = -\dfrac{C}{(\ln 2)^3} \implies C = 0$$

Then, we write the particular solution.

$$x = -\dfrac{1}{2t} = -\dfrac{1}{2\ln(\cos y)}$$

Continuing, we find $$x\!\left(\dfrac{\pi}{6}\right)$$.

$$\cos\!\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}$$

$$\ln\!\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{1}{2}\ln 3 - \ln 2 = \dfrac{\ln 3 - 2\ln 2}{2}$$

$$x\!\left(\dfrac{\pi}{6}\right) = -\dfrac{1}{2 \cdot \frac{\ln 3 - 2\ln 2}{2}} = -\dfrac{1}{\ln 3 - 2\ln 2} = \dfrac{1}{2\ln 2 - \ln 3} = \dfrac{1}{\ln 4 - \ln 3}$$

It follows that we identify $$m$$ and $$n$$.

Comparing with $$\dfrac{1}{\log_e m - \log_e n}$$:

$$m = 4, \quad n = 3$$

Since $$\gcd(4, 3) = 1$$, they are coprime.

$$mn = 4 \times 3 = \boxed{12}$$