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