Step 1: Differentiate the parametric equations.
With $x=\sin2\theta+\sin3\theta$ and $y=\cos2\theta-\cos3\theta$, \[ \frac{dx}{d\theta}=2\cos2\theta+3\cos3\theta,\qquad \frac{dy}{d\theta}=-2\sin2\theta+3\sin3\theta. \]
Step 2: First derivative.
\[ \frac{dy}{dx}=\frac{-2\sin2\theta+3\sin3\theta}{2\cos2\theta+3\cos3\theta}. \]
Step 3: Set up the second derivative.
For parametric curves, \[ \frac{d^2y}{dx^2}=\frac{\dfrac{d}{d\theta}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{d\theta}}. \]
Step 4: Differentiate the ratio.
Apply the quotient rule to $\dfrac{dy}{dx}$. The numerator of $\dfrac{d}{d\theta}\!\left(\dfrac{dy}{dx}\right)$ collapses, using product-to-sum identities, to $19+6\cos\theta$, over $(2\cos2\theta+3\cos3\theta)^2$.
Step 5: Divide by dx/dθ.
Dividing once more by $\dfrac{dx}{d\theta}=2\cos2\theta+3\cos3\theta$ raises the denominator power to $3$: \[ \frac{d^2y}{dx^2}=\frac{19+6\cos\theta}{(2\cos2\theta+3\cos3\theta)^3}. \]
Step 6: Box the answer.
This matches option (2).
\[ \boxed{\frac{d^2y}{dx^2}=\frac{19+6\cos\theta}{(2\cos2\theta+3\cos3\theta)^3}\ \text{(option 2)}} \]