If for \(x \in\left(0, \frac{1}{4}\right)\), the derivative of \(\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}\right)\) is \(\sqrt{x} \cdot g(x)\), then \(g(x)\) equals

Let \(y=\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}\right)\)
\(\begin{aligned}
& \quad=\tan ^{-1}\left(\frac{3 x \sqrt{x}+3 x \sqrt{x}}{1-3 x \sqrt{x} \cdot 3 x \sqrt{x}}\right) \\
& \quad=\tan ^{-1}(3 x \sqrt{x})+\tan ^{-1} 3 x \sqrt{x} \\
& y=2 \tan ^{-1}(3 x \cdot \sqrt{x})
\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{2}{1+(3 x \sqrt{x})^2} \cdot \frac{\mathrm{~d}}{\mathrm{~d} x}(3 x \cdot \sqrt{x}) \\
& =\frac{2}{1+9 x^3} \cdot \frac{9}{2} \sqrt{x} \\
\frac{\mathrm{~d} y}{\mathrm{~d} x} & =\frac{9 \sqrt{x}}{1+9 x^3}
\end{aligned}\)
\(\therefore \quad\) Comparing with \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\sqrt{x} \cdot \mathrm{~g}(x)\)
\(\therefore \quad \mathrm{g}(x)=\frac{9}{1+9 x^3}\)