Mathematics Question from JEE Main 2023 · 24 January, Shift 2

Let $f$ be a differentiable function defined on $\left(0, \frac{\pi}{2}\right)$ such that $f(x) > 0$ and

$$f(x) + \int_0^x f(t)\sqrt{1 - (\log_e f(t))^2} \, dt = e, \quad \forall x \in \left[0, \frac{\pi}{2}\right].$$

Then $\left(6 \log_e f\left(\frac{\pi}{6}\right)\right)^2$ is equal to: