${\tan ^{ - 1}}\left( {\frac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right),\frac{{ - 1}}{{\sqrt 3 }} < \frac{x}{a} < \frac{1}{{\sqrt 3 }}$

Let $y = {\tan ^{ - 1}}\left( {\frac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$

Put $x = {\mathop{\rm a}\nolimits} \,tan\theta \Rightarrow \theta = {\tan ^{ - 1}}\frac{x}{a}$

therefore,$y = {\tan ^{ - 1}}\left[ {\frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right]$

$= {\tan ^{ - 1}}(\tan 3\theta ) = 3\theta$
$= 3{\tan ^{ - 1}}\frac{x}{a}$

therefore,$\frac{{dy}}{{dx}} = 3 \cdot \frac{d}{{dx}}{\tan ^{ - 1}}\frac{x}{a} = 3 \cdot \left[ {\frac{1}{{1 + \frac{{{x^2}}}{{{a^2}}}}}} \right] \cdot \frac{d}{{dx}} \cdot \left( {\frac{x}{a}} \right)$

$= 3 \cdot \frac{{{a^2}}}{{{a^2} + {x^2}}} \cdot \frac{1}{a} = \frac{{3a}}{{{a^2} + {x^2}}}$

$\frac{dy}{dx} = \frac{{3a}}{{{a^2} + {x^2}}}$