If $y = \tan^{-1}\left[\dfrac{x - \sqrt{1 - x^2}}{x + \sqrt{1 - x^2}}\right]$ , then $\dfrac{dy}{dx} =$

$\cos\left(\cos^{-1}\left(-\dfrac{1}{2}\right) + \dfrac{\pi}{3}\right) =$

$\sin^{-1}\left(\dfrac{12}{13}\right) + \cos^{-1}\left(\dfrac{4}{5}\right) + \tan^{-1}\left(\dfrac{63}{16}\right) =$

If $f(x) = \dfrac{4x + 3}{6x - 4}$, $x \neq \dfrac{2}{3}$ and $(\text{fof})(x) = g(x)$ where $g: \mathbb{R} - \left\{\dfrac{2}{3}\right\} \rightarrow \mathbb{R} - \left\{\dfrac{2}{3}\right\}$, then $(g\,o\,g\,o\,g\,o\,g\,o\,g)\,(3) =$