Let $$y = y(x)$$ be the solution of the differential equation $$\sec^2 x \, dx + e^{2y}(\tan^2 x + \tan x) \, dy = 0$$, $$0 \lt x \lt \frac{\pi}{2}$$, $$y\left(\frac{\pi}{4}\right) = 0$$. If $$y\left(\frac{\pi}{6}\right) = \alpha$$, then $$e^{8\alpha}$$ is equal to

We are given the differential equation $$\sec^2 x\,dx + e^{2y}(\tan^2 x + \tan x)\,dy = 0$$ with $$y\!\left(\frac{\pi}{4}\right) = 0$$, and we need to find $$e^{8\alpha}$$ where $$y\!\left(\frac{\pi}{6}\right) = \alpha$$.

Writing the equation as $$\frac{dx}{dy} = -\frac{e^{2y}(\tan^2 x + \tan x)}{\sec^2 x}$$, we substitute $$v = \tan x$$ so that $$\frac{dv}{dy} = \sec^2 x \cdot \frac{dx}{dy}$$. This gives:

$$\frac{dv}{dy} = -(e^{2y}\,v^2 + v).$$

Rearranging: $$\frac{dv}{dy} + v = -e^{2y}\,v^2$$. This is a Bernoulli equation with $$n = 2$$.

Dividing both sides by $$v^2$$: $$v^{-2}\frac{dv}{dy} + v^{-1} = -e^{2y}$$.

Substituting $$w = v^{-1} = \cot x$$, so $$\frac{dw}{dy} = -v^{-2}\frac{dv}{dy}$$, we obtain:

$$-\frac{dw}{dy} + w = -e^{2y} \quad\Longrightarrow\quad \frac{dw}{dy} - w = e^{2y}.$$

This is a first-order linear ODE. The integrating factor is $$\mu = e^{\int -1\,dy} = e^{-y}$$. Multiplying through:

$$\frac{d}{dy}\!\left(w\,e^{-y}\right) = e^{2y} \cdot e^{-y} = e^{y}.$$

Integrating both sides: $$w\,e^{-y} = e^{y} + C$$, hence $$w = e^{2y} + C\,e^{y}$$.

Since $$w = \cot x$$, we have $$\cot x = e^{2y} + C\,e^{y}$$.

Applying the initial condition $$x = \frac{\pi}{4},\; y = 0$$: $$\cot\frac{\pi}{4} = 1 = e^{0} + C\,e^{0} = 1 + C$$, giving $$C = 0$$.

The particular solution is $$\cot x = e^{2y}$$.

Now substituting $$x = \frac{\pi}{6},\; y = \alpha$$: $$\cot\frac{\pi}{6} = \sqrt{3} = e^{2\alpha}$$.

Therefore $$e^{2\alpha} = 3^{1/2}$$, and raising both sides to the fourth power:

$$e^{8\alpha} = \left(e^{2\alpha}\right)^4 = \left(\sqrt{3}\right)^4 = 9.$$

So, the answer is $$9$$.