`x=3costheta-2cos^(3)theta, y = 3 sin theta- 2 sin^(3) theta`

Explanation

To solve the problem, we need to differentiate the parametric equations given by:

\[ x = 3 \cos \theta - 2 \cos^3 \theta \]
\[ y = 3 \sin \theta - 2 \sin^3 \theta \]

We will find \(\frac{dy}{dx}\) using the chain rule.

### Step 1: Differentiate \(x\) with respect to \(\theta\)

We differentiate \(x\) as follows:

\[ \frac{dx}{d\theta} = \frac{d}{d\theta}(3 \cos \theta) - \frac{d}{d\theta}(2 \cos^3 \theta) \]

Using the chain rule and product rule:

\[ \frac{dx}{d\theta} = -3 \sin \theta - 2 \cdot 3 \cos^2 \theta \cdot (-\sin \theta) \]
\[ = -3 \sin \theta + 6 \cos^2 \theta \sin \theta \]
\[ = \sin \theta (6 \cos^2 \theta - 3) \]

### Step 2: Differentiate \(y\) with respect to \(\theta\)

Now we differentiate \(y\):

\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(3 \sin \theta) - \frac{d}{d\theta}(2 \sin^3 \theta) \]

Again using the chain rule and product rule:

\[ \frac{dy}{d\theta} = 3 \cos \theta - 2 \cdot 3 \sin^2 \theta \cdot \cos \theta \]
\[ = 3 \cos \theta - 6 \sin^2 \theta \cos \theta \]
\[ = \cos \theta (3 - 6 \sin^2 \theta) \]

### Step 3: Find \(\frac{dy}{dx}\)

Now we can find \(\frac{dy}{dx}\) using the formula:

\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \]

Substituting the derivatives we found:

\[ \frac{dy}{dx} = \frac{\cos \theta (3 - 6 \sin^2 \theta)}{\sin \theta (6 \cos^2 \theta - 3)} \]

### Step 4: Simplify the expression

Now we can simplify the expression:

\[ \frac{dy}{dx} = \frac{\cos \theta (3 - 6 \sin^2 \theta)}{\sin \theta (6 \cos^2 \theta - 3)} \]

Using the identities \(3 - 6 \sin^2 \theta = 3(1 - 2 \sin^2 \theta) = 3 \cos 2\theta\) and \(6 \cos^2 \theta - 3 = 3(2 \cos^2 \theta - 1) = 3 \cos 2\theta\):

\[ \frac{dy}{dx} = \frac{\cos \theta \cdot 3 \cos 2\theta}{\sin \theta \cdot 3 \cos 2\theta} \]

Cancelling \(3 \cos 2\theta\) (assuming \(\cos 2\theta \neq 0\)):

\[ \frac{dy}{dx} = \frac{\cos \theta}{\sin \theta} = \cot \theta \]

### Final Answer

\[ \frac{dy}{dx} = \cot \theta \]