Step-by-step parametric differentiation:
Given $x = a\sin^3 t$ and $y = b\cos^3 t$:

• First derivatives with respect to $t$:
  $\frac{dx}{dt} = 3a\sin^2 t \cos t$, and $\frac{dy}{dt} = -3b\cos^2 t \sin t$.
• First derivative $\frac{dy}{dx}$:
  $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-3b\cos^2 t \sin t}{3a\sin^2 t \cos t} = -\frac{b}{a} \cot t$.
• Second derivative $\frac{d^2 y}{dx^2}$:
  $\frac{d^2 y}{dx^2} = \frac{d}{dt}\left(-\frac{b}{a}\cot t\right) \cdot \frac{dt}{dx} = \left(\frac{b}{a} \operatorname{cosec}^2 t\right) \cdot \frac{1}{3a\sin^2 t \cos t}$
  $= \frac{b}{3a^2} \frac{\operatorname{cosec}^2 t}{\sin^2 t \cos t} = \frac{b}{3a^2} \operatorname{cosec}^4 t \sec t$.
  
  Option breakdown:
• Inverts the trigonometric functions into numerator powers.
• $\frac{b}{3a^2}\operatorname{cosec}^4 t \sec t$ is the exact second derivative.
• Inverts the constant coefficient to $a/(3b^2)$.
• Swaps the coefficients $a$ and $b$.