If for x ∈ (0, (1)/(4)), the derivative of ^ - 1 ( 6x x 1 - 9 x ^ 3...

Let $3{x}^{\dfrac{3}{2}} = y$, then $y \in \left(0, \dfrac{3}{8}\right)$

$\tan^{- 1} \left(\dfrac{6x\sqrt{x}}{1 - 9{x}^{3}}\right) = \tan^{- 1} \left(\dfrac{2y}{1 - {y}^{2}}\right)$

$⇒ 2\tan^{- 1} y = 2\tan^{- 1} \left(3x\dfrac{3}{2}\right)$

Now $⇒ \dfrac{d}{dx}2\tan^{- 1} \left(3x\dfrac{3}{2}\right) = \dfrac{2}{1 + 9{x}^{3}} ⋅ 3 ⋅ \dfrac{3}{2} ⋅ x\dfrac{1}{2}$

$⇒ \dfrac{d}{dx}2\tan^{- 1} \left(3x\dfrac{3}{2}\right) = \dfrac{9}{1 + 9{x}^{3}} ⋅ \sqrt{x}$

$⇒ \sqrt{x} . g\left(x\right) = \dfrac{9}{1 + 9{x}^{3}} ⋅ \sqrt{x}$

$⇒ g\left(x\right) = \dfrac{9}{1 + 9{x}^{3}}$