Let $$f$$ be a differentiable function defined on $$\left[0, \frac{\pi}{2}\right]$$ such that $$f(x) > 0$$ and $$f(x) + \int_0^x f(t)\sqrt{1 - (\log_e(f(t)))^2} dt = e$$ $$\forall x \in \left[0, \frac{\pi}{2}\right]$$, then $$\left\{6\log_e\left(f\left(\frac{\pi}{6}\right)\right)\right\}^2$$ is equal to

We are given a differentiable function $$f$$ on $$\left[0, \dfrac{\pi}{2}\right]$$ with $$f(x) > 0$$ satisfying:

$$f(x) + \int_0^x f(t)\sqrt{1 - (\log_e f(t))^2}\,dt = e$$

Differentiate both sides with respect to $$x$$ to obtain

$$f'(x) + f(x)\sqrt{1 - (\ln f(x))^2} = 0$$

Let $$u = \ln f(x)$$. Since $$\dfrac{du}{dx} = \dfrac{f'(x)}{f(x)}$$, the differential equation becomes

$$\dfrac{du}{dx} = -\sqrt{1 - u^2}$$

This equation is separable, so

$$\dfrac{du}{\sqrt{1 - u^2}} = -dx$$

Integrating gives

$$\arcsin(u) = -x + C$$

At $$x = 0$$, we have $$f(0) + 0 = e$$, hence $$f(0) = e$$ and $$u(0) = \ln e = 1$$.

Since $$\arcsin(1) = C$$, it follows that $$C = \dfrac{\pi}{2}$$ and therefore

$$\ln f(x) = \sin\!\Bigl(\dfrac{\pi}{2} - x\Bigr) = \cos x$$

Evaluating at $$x = \dfrac{\pi}{6}$$ yields $$\ln f\!\Bigl(\dfrac{\pi}{6}\Bigr) = \cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$$

Thus

$$\Bigl\{6\log_e f\!\Bigl(\dfrac{\pi}{6}\Bigr)\Bigr\}^2 = \Bigl(6 \cdot \dfrac{\sqrt{3}}{2}\Bigr)^2 = (3\sqrt{3})^2 = 27$$

The answer is $$\boxed{27}$$.