The provided parametric equations are:\[x = a(\cos\theta + \theta\sin\theta), \quad y = a(\sin\theta - \theta\cos\theta).\]Step 1: Compute \(\frac{dx}{d\theta}\). By differentiating \( x \) with respect to \(\theta\), we get:\[\frac{dx}{d\theta} = a\left(-\sin\theta + \theta\cos\theta + \sin\theta + \theta\cos\theta\right).\]After simplification:\[\frac{dx}{d\theta} = a\theta\cos\theta.\]Step 2: Compute \(\frac{dy}{d\theta}\). Differentiating \( y \) with respect to \(\theta\) yields:\[\frac{dy}{d\theta} = a\left(\cos\theta + \theta(-\sin\theta) - \cos\theta - \theta(-\sin\theta)\right).\]Upon simplification:\[\frac{dy}{d\theta} = a\theta\sin\theta.\]Step 3: Compute \(\frac{dy}{dx}\). Applying the chain rule, we have:\[\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}.\]Substituting the computed derivatives:\[\frac{dy}{dx} = \frac{a\theta\sin\theta}{a\theta\cos\theta} = \tan\theta.\]Step 4: Compute \(\frac{d^2y}{dx^2}\). We differentiate \(\frac{dy}{dx} = \tan\theta\) with respect to \( x \):\[\frac{d^2y}{dx^2} = \frac{d}{dx}(\tan\theta).\]Using the chain rule again:\[\frac{d^2y}{dx^2} = \sec^2\theta \cdot \frac{d\theta}{dx}.\]From Step 1, we derived:\[\frac{dx}{d\theta} = a\theta\cos\theta, \quad \text{which implies} \quad \frac{d\theta}{dx} = \frac{1}{a\theta\cos\theta}.\]Substituting this into the expression for \(\frac{d^2y}{dx^2}\):\[\frac{d^2y}{dx^2} = \sec^2\theta \cdot \frac{1}{a\theta\cos\theta}.\]Simplifying the expression results in:\[\frac{d^2y}{dx^2} = \frac{\sec^3\theta}{a\theta}.\] Final Answer:\[\boxed{\frac{\sec^3\theta}{a\theta} }.\]