1. Let $y=\tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right).$

We need to differentiate this and express it in the form $\frac{dy}{dx}=\sqrt{x}\,g(x).$

2. First simplify the argument: $6x\sqrt{x}=6x^{3/2}.$ So, $y=\tan^{-1}\left(\frac{6x^{3/2}}{1-9x^3}\right).$

3. Notice the standard identity: $\tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta}.$

Take $\tan\theta=3x^{3/2}.$ Then $\tan 2\theta=\frac{2(3x^{3/2})}{1-(3x^{3/2})^2} = \frac{6x^{3/2}}{1-9x^3}.$

Hence, $y=\tan^{-1}(\tan 2\theta).$

Since $x\in(0,1/4)$, we have $3x^{3/2}>0$ and small enough so that the principal value issue does not arise. Therefore, $y=2\tan^{-1}(3x^{3/2}).$

4. Differentiate: $\frac{dy}{dx}=2\cdot \frac{1}{1+(3x^{3/2})^2}\cdot \frac{d}{dx}(3x^{3/2}).$

Now, $\frac{d}{dx}(3x^{3/2})=3\cdot \frac{3}{2}x^{1/2}=\frac{9}{2}\sqrt{x}.$

Thus,