To find $\frac{dy}{dx}$, we can use parametric differentiation, as both $x$ and $y$ are given as functions of $t$.

The equations are: $x = a \cos^3 t, \quad y = a \sin^3 t$

Steps:

1. Compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$: $\frac{dx}{dt} = a \cdot 3 \cos^2 t (-\sin t) = -3a \cos^2 t \sin t$ $\frac{dy}{dt} = a \cdot 3 \sin^2 t (\cos t) = 3a \sin^2 t \cos t$

2. Use the chain rule for parametric differentiation: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

   Substituting the values: $\frac{dy}{dx} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t}$

3. Simplify: $\frac{dy}{dx} = \frac{\sin^2 t \cos t}{-\cos^2 t \sin t} = -\frac{\sin t}{\cos t} = -\tan t$

Final Answer:

$\frac{dy}{dx} = -\tan t$

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