Q.$x=a\cos\theta$, $y=b\sin\theta$, at $\theta=\pi/4$

Given $x=a\cos\theta$, $y=b\sin\theta$.

Step 1: $\dfrac{dx}{d\theta}=-a\sin\theta$, $\dfrac{dy}{d\theta}=b\cos\theta$, so $\dfrac{dy}{dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac{b}{a}\cot\theta$.

Step 2 — differentiate w.r.t. $\theta$: $\dfrac{d}{d\theta}\left(-\dfrac{b}{a}\cot\theta\right)=\dfrac{b}{a}\csc^2\theta$.

Step 3 — divide by $dx/d\theta=-a\sin\theta$: $\dfrac{d^2y}{dx^2}=\dfrac{\frac{b}{a}\csc^2\theta}{-a\sin\theta}=-\dfrac{b}{a^2}\cdot\dfrac{\csc^2\theta}{\sin\theta}=-\dfrac{b}{a^2}\csc^3\theta$. …

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