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

Question

$\begin{array}{1 1} (1) \large\frac{3 x \sqrt x}{1-9x^3} \\ (2) \large\frac{3 x }{1-9x^3} \\ (1) \large\frac{3}{1+9x^3} \\ (4) \large\frac{9}{1+9x^3} \end{array} $