Step 1: Understand parametric form.
Here both $x$ and $y$ are written using a helper variable $t$. To get $\frac{dy}{dx}$ we cannot differentiate directly. Instead we differentiate each with respect to $t$ and divide.
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Step 2: Differentiate $x$ with respect to $t$.
We have $x = a\cos^3 t$. Using the chain rule, bring down the power $3$ and differentiate $\cos t$.
\[ \frac{dx}{dt} = 3a\cos^2 t\,(-\sin t) = -3a\cos^2 t\,\sin t \]
Step 3: Differentiate $y$ with respect to $t$.
We have $y = a\sin^3 t$. Again bring down the power and differentiate $\sin t$.
\[ \frac{dy}{dt} = 3a\sin^2 t\,(\cos t) = 3a\sin^2 t\,\cos t \]
Step 4: Divide to get $\frac{dy}{dx}$.
Put the two results over each other. Many factors cancel.
\[ \frac{dy}{dx} = \frac{3a\sin^2 t\,\cos t}{-3a\cos^2 t\,\sin t} = -\frac{\sin t}{\cos t} = -\tan t \]
Step 5: Plug in the value of $t$.
We need the slope at $t = \frac{\pi}{4}$. We know $\tan\frac{\pi}{4} = 1$.
\[ \frac{dy}{dx} = -\tan\frac{\pi}{4} = -1 \]
Step 6: State the answer.
So the slope of the curve at that point is $-1$.
\[ \boxed{-1} \]