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

Let us denote the given function by

$$y \;=\; \tan^{-1}\!\!\left(\dfrac{6x\sqrt{x}}{1-9x^{3}}\right),\qquad 0<x<\dfrac14.$$

To find $$\dfrac{dy}{dx}$$ we recall the standard formula

$$\dfrac{d}{dx}\bigl(\tan^{-1}t\bigr)\;=\;\dfrac{t'}{1+t^{2}},$$

where $$t=t(x).$$ Here

$$t(x)=\dfrac{6x\sqrt{x}}{1-9x^{3}}=\dfrac{6x^{3/2}}{1-9x^{3}}.$$

We first differentiate $$t(x)$$. Writing $$t(x)=\dfrac{u}{v}$$ with

$$u=6x^{3/2},\qquad v=1-9x^{3},$$

we use the quotient rule $$\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}}.$$

Compute $$u'$$:

$$u'=6\cdot\dfrac{3}{2}x^{1/2}=9x^{1/2}.$$
Compute $$v'$$:

$$v'=-27x^{2}.$$

Hence

$$ \begin{aligned} t'(x)&=\dfrac{u'v-uv'}{v^{2}}\\[4pt] &=\dfrac{9x^{1/2}(1-9x^{3})-6x^{3/2}(-27x^{2})}{(1-9x^{3})^{2}}\\[4pt] &=\dfrac{9x^{1/2}-81x^{7/2}+162x^{7/2}}{(1-9x^{3})^{2}}\\[4pt] &=\dfrac{9x^{1/2}+81x^{7/2}}{(1-9x^{3})^{2}}\\[4pt] &=\dfrac{9x^{1/2}\bigl(1+9x^{3}\bigr)}{(1-9x^{3})^{2}}. \end{aligned} $$

Now we need $$1+t^{2}$$. Since

$$t(x)=\dfrac{6x^{3/2}}{1-9x^{3}},\quad t^{2}(x)=\dfrac{36x^{3}}{(1-9x^{3})^{2}},$$

we get

$$ 1+t^{2}=1+\dfrac{36x^{3}}{(1-9x^{3})^{2}} =\dfrac{(1-9x^{3})^{2}+36x^{3}}{(1-9x^{3})^{2}}. $$

Observe that

$$ (1-9x^{3})^{2}+36x^{3}=1-18x^{3}+81x^{6}+36x^{3}=1+18x^{3}+81x^{6}=(1+9x^{3})^{2}. $$

Therefore

$$1+t^{2}=\dfrac{(1+9x^{3})^{2}}{(1-9x^{3})^{2}}.$$

Putting everything into the derivative formula, we have

$$ \begin{aligned} \dfrac{dy}{dx} &=\dfrac{t'}{1+t^{2}} =\dfrac{\displaystyle \dfrac{9x^{1/2}(1+9x^{3})}{(1-9x^{3})^{2}}} {\displaystyle \dfrac{(1+9x^{3})^{2}}{(1-9x^{3})^{2}}}\\[8pt] &=\dfrac{9x^{1/2}(1+9x^{3})}{(1-9x^{3})^{2}}\cdot \dfrac{(1-9x^{3})^{2}}{(1+9x^{3})^{2}}\\[6pt] &=\dfrac{9x^{1/2}}{1+9x^{3}}. \end{aligned} $$

Thus

$$\dfrac{dy}{dx}= \sqrt{x}\;\cdot\;\dfrac{9}{1+9x^{3}}.$$

Comparing with the given form $$\dfrac{dy}{dx}= \sqrt{x}\,g(x),$$ we finally obtain

$$g(x)=\dfrac{9}{1+9x^{3}}.$$

Hence, the correct answer is Option A.