If $x = a\sec^3 \theta$, $y = a \tan^3 \theta$, then $\frac{dy}{dx}$ at $\theta = \frac{\pi}{3}$ is

Given $x = a\sec^3\theta$ and $y = a\tan^3\theta$.

Since both $x$ and $y$ are in terms of $\theta$, use parametric differentiation:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

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For $x = a\sec^3\theta$:

$\frac{dx}{d\theta} = 3a\sec^2\theta \cdot \sec\theta\tan\theta$

$\frac{dx}{d\theta} = 3a\sec^3\theta\tan\theta$

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For $y = a\tan^3\theta$:

$\frac{dy}{d\theta} = 3a\tan^2\theta \cdot \sec^2\theta$

$\frac{dy}{d\theta} = 3a\tan^2\theta\sec^2\theta$

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$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

$\frac{dy}{dx} = \frac{3a\tan^2\theta\sec^2\theta}{3a\sec^3\theta\tan\theta}$

$\frac{dy}{dx} = \frac{\tan^2\theta\sec^2\theta}{\sec^3\theta\tan\theta}$

$\frac{dy}{dx} = \frac{\tan\theta}{\sec\theta}$

$\frac{dy}{dx} = \frac{\sin\theta/\cos\theta}{1/\cos\theta}$

$\frac{dy}{dx} = \sin\theta$

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At $\theta = \frac{\pi}{3}$:

$\frac{dy}{dx}\bigg|_{\theta=\pi/3} = \sin\left(\frac{\pi}{3}\right)$

$\frac{dy}{dx}\bigg|_{\theta=\pi/3} = \frac{\sqrt{3}}{2}$