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Given $x = a \sin^3 \theta$ and $y = b \cos^3 \theta$. We need to find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$. $$\frac{dx}{d\theta} = \frac{d}{d\theta}(a \sin^3 \theta) = 3a \sin^2 \theta \cos \theta$$ $$\frac{dy}{d\theta} = \frac{d}{d\theta}(b \cos^3 \theta) = -3b \cos^2 \theta \sin \theta$$
Now, we find $\frac{dy}{dx}$ using the chain rule: $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-3b \cos^2 \theta \sin \theta}{3a \sin^2 \theta \cos \theta} = -\frac{b}{a} \frac{\cos \theta}{\sin \theta} = -\frac{b}{a} \cot \theta$$
Next, we find $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ with respect to $x$: $$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} \left( -\frac{b}{a} \cot \theta \right) = -\frac{b}{a} \frac{d}{d\theta} (\cot \theta) \cdot \frac{d\theta}{dx}$$ We know that $\frac{d}{d\theta} (\cot \theta) = -\csc^2 \theta$ and $\frac{dx}{d\theta} = 3a \sin^2 \theta \cos \theta$, so $\frac{d\theta}{dx} = \frac{1}{3a \sin^2 \theta \cos \theta}$. Therefore, $$\frac{d^2y}{dx^2} = -\frac{b}{a} (-\csc^2 \theta) \cdot \frac{1}{3a \sin^2 \theta \cos \theta} = \frac{b}{3a^2} \frac{1}{\sin^4 \theta \cos \theta}$$
Now, we evaluate $\frac{d^2y}{dx^2}$ at $\theta = \frac{\pi}{4}$. At $\theta = \frac{\pi}{4}$, $\sin \theta = \frac{1}{\sqrt{2}}$ and $\cos \theta = \frac{1}{\sqrt{2}}$. $$\frac{d^2y}{dx^2} \Big|_{\theta = \frac{\pi}{4}} = \frac{b}{3a^2} \frac{1}{(\frac{1}{\sqrt{2}})^4 (\frac{1}{\sqrt{2}})} = \frac{b}{3a^2} \frac{1}{\frac{1}{4} \cdot \frac{1}{\sqrt{2}}} = \frac{b}{3a^2} \cdot 4\sqrt{2} = \frac{4\sqrt{2}b}{3a^2}$$
Final Answer: $\frac{4\sqrt{2}b}{3a^2}$
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