If $y={\mathrm{Tan}}^{-1}\left(\frac{3a^{2}x-x^3}{a^3-3a{x}^2}\right)\text{then}\frac{dy}{dx}=$

$\begin{array}{rl}y& ={\tan }^{-1}\frac{3a^{2}x-x^3}{a\left(a^2-3x^2\right)}\end{array}$

$={\tan }^{-1}\frac{3(x/a)-(x/a{)}^3}{1-3(x/a{)}^2}$

$\text{ (putting }x/a=\tan \theta )$

$={\tan }^{-1}\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$

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

$\begin{array}{rl}\therefore \frac{dy}{dx}& =3\frac{d}{dx}{\tan }^{-1}(x/a)\end{array}$

$=3\frac{1}{1+(x/a{)}^2}\times \frac{1}{a}=\frac{3a}{a^2+x^2}$