Limits

Let $y = \tan^{-1}\left(\dfrac {6x\sqrt x}{1-9x^3} \right)$ , then we have:

$\begin{aligned} \tan y & = \frac {6x\sqrt x}{1-9x^3} \\ & = \frac {2{\color{#3D99F6}(3x\sqrt x)}}{1-{\color{#3D99F6}(3x\sqrt x)}^2} \\ \implies \tan \frac y2 & = 3x\sqrt x & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ {\color{#3D99F6}\sec^2 \frac y2} \frac {dy}{dx} & = \frac 92 \sqrt x \\ \implies \frac {dy}{dx} & = \frac {9\sqrt x}{\color{#3D99F6} 1+\tan^2 \frac y2} \\ & = \frac {9\sqrt x}{1+(3x\sqrt x)^2} \\ & = \sqrt x \cdot \frac 9{1+9x^3} \\ \implies g(x) & = \boxed{\dfrac 9{1+9x^3}} \end{aligned}$