If $\mathrm{y}=\tan ^{-1}\left(\frac{12 x-64 x^3}{1-48 x^2}\right)$, then $\frac{\mathrm{dy}}{\mathrm{d} x}=$

Given $y=\tan^{-1}\left(\frac{12x-64x^3}{1-48x^2}\right)$, we recognize the expression inside the inverse tangent resembles the triple angle formula for tangent: $\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$.

Let $4x = \tan\theta$, so $x = \frac{1}{4}\tan\theta$. Substituting this into $y$ gives:

$y = \tan^{-1}\left(\frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}\right)$.

This simplifies to $y = \tan^{-1}(\tan(3\theta)) = 3\theta$, where $\theta = \tan^{-1}(4x)$.

Thus, $y = 3\tan^{-1}(4x)$.

Differentiating using the chain rule:

$\frac{dy}{dx} = 3 \cdot \frac{1}{1+(4x)^2} \cdot 4 = \frac{12}{1+16x^2}$.

The result matches option C.