If $x = a\sec^3 \theta$, $y = a \tan^3 \theta$, then $\frac{d^2y}{dx^2}$ equals.

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

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

$= 3a\sec^3\theta\tan\theta$

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

$= 3a\tan^2\theta\sec^2\theta$

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

$= \dfrac{3a\tan^2\theta\sec^2\theta}{3a\sec^3\theta\tan\theta}$

$= \dfrac{\tan\theta}{\sec\theta}$

$= \tan\theta \cdot \cos\theta$

$= \dfrac{\sin\theta}{\cos\theta}\cdot\cos\theta$

$= \sin\theta$

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$\dfrac{d^2y}{dx^2} = \dfrac{\dfrac{d}{d\theta}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{d\theta}}$

$\dfrac{d}{d\theta}(\sin\theta) = \cos\theta$

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

$= 3a\cdot\dfrac{1}{\cos^3\theta}\cdot\dfrac{\sin\theta}{\cos\theta}$

$= \dfrac{3a\sin\theta}{\cos^4\theta}$

$\dfrac{d^2y}{dx^2} = \dfrac{\cos\theta}{\;\dfrac{3a\sin\theta}{\cos^4\theta}\;}$

$= \cos\theta \times \dfrac{\cos^4\theta}{3a\sin\theta}$

$= \dfrac{\cos^5\theta}{3a\sin\theta}$