If $x = a \sec^3 \theta$, $y = a \tan^3 \theta$, find $\dfrac{d^2 y}{dx^2}$ at $\theta = \dfrac{\pi}{3}$
$x = a \sec^3 \theta$
$\therefore$ $\dfrac{dx}{d \theta} = 3 a \sec^2 \theta \sec \theta \tan \theta = 3 a \sec^3 \theta \tan \theta$
$y = a \tan^3 \theta$
$\therefore$ $\dfrac{dy}{d \theta} = 3 a \tan^2 \theta \sec^2 \theta$
$\therefore$ $\dfrac{dy}{dx} = \dfrac{dy/d \theta}{dx / d \theta} = \dfrac{3 a \tan^2 \theta \sec^2 \theta}{3 a \sec^3 \theta \tan \theta} = \dfrac{\tan \theta}{\sec \theta}$
i.e. $\dfrac{dy}{dx} = \dfrac{\sin \theta / \cos \theta}{1 / \cos \theta} = \sin \theta$
$\therefore$ $\dfrac{d^2 y}{dx^2} = \cos \theta \times \dfrac{d \theta}{dx} = \dfrac{\cos \theta}{dx / d \theta} = \dfrac{\cos \theta}{3 a \sec^3 \theta \tan \theta}$
i.e. $\dfrac{d^2 y}{dx^2}$ $\bigg|_{\theta = \frac{\pi}{3}}$ $= \dfrac{\cos \left(\pi / 3\right)}{3a \sec^3 \left(\pi / 3\right) \tan \left(\pi / 3\right)} = \dfrac{1/2}{3a \times 2^3 \times \sqrt{3}} = \dfrac{1}{48a\sqrt{3}}$