`x=e^(theta)(theta+(1)/(theta)),y = e^(-theta)(theta-(1)/(theta))`

Explanation

To find \(\frac{dy}{dx}\) given the parametric equations \(x = e^{\theta} \left( \theta + \frac{1}{\theta} \right)\) and \(y = e^{-\theta} \left( \theta - \frac{1}{\theta} \right)\), we will follow these steps:

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

Given:
\[ x = e^{\theta} \left( \theta + \frac{1}{\theta} \right) \]

Using the product rule:
\[ \frac{dx}{d\theta} = \frac{d}{d\theta} \left( e^{\theta} \right) \left( \theta + \frac{1}{\theta} \right) + e^{\theta} \frac{d}{d\theta} \left( \theta + \frac{1}{\theta} \right) \]

Calculating the derivatives:
1. \(\frac{d}{d\theta} e^{\theta} = e^{\theta}\)
2. \(\frac{d}{d\theta} \left( \theta + \frac{1}{\theta} \right) = 1 - \frac{1}{\theta^2}\)

So,
\[ \frac{dx}{d\theta} = e^{\theta} \left( \theta + \frac{1}{\theta} \right) + e^{\theta} \left( 1 - \frac{1}{\theta^2} \right) \]

Combining terms:
\[ \frac{dx}{d\theta} = e^{\theta} \left( \theta + \frac{1}{\theta} + 1 - \frac{1}{\theta^2} \right) \]

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

Given:
\[ y = e^{-\theta} \left( \theta - \frac{1}{\theta} \right) \]

Using the product rule:
\[ \frac{dy}{d\theta} = \frac{d}{d\theta} \left( e^{-\theta} \right) \left( \theta - \frac{1}{\theta} \right) + e^{-\theta} \frac{d}{d\theta} \left( \theta - \frac{1}{\theta} \right) \]

Calculating the derivatives:
1. \(\frac{d}{d\theta} e^{-\theta} = -e^{-\theta}\)
2. \(\frac{d}{d\theta} \left( \theta - \frac{1}{\theta} \right) = 1 + \frac{1}{\theta^2}\)

So,
\[ \frac{dy}{d\theta} = -e^{-\theta} \left( \theta - \frac{1}{\theta} \right) + e^{-\theta} \left( 1 + \frac{1}{\theta^2} \right) \]

Combining terms:
\[ \frac{dy}{d\theta} = e^{-\theta} \left( -\theta + \frac{1}{\theta} + 1 + \frac{1}{\theta^2} \right) \]

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

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]

Substituting the expressions we found:
\[ \frac{dy}{dx} = \frac{e^{-\theta} \left( -\theta + \frac{1}{\theta} + 1 + \frac{1}{\theta^2} \right)}{e^{\theta} \left( \theta + \frac{1}{\theta} + 1 - \frac{1}{\theta^2} \right)} \]

### Step 4: Simplify the expression

The \(e^{-\theta}\) in the numerator and \(e^{\theta}\) in the denominator combine to give:
\[ \frac{dy}{dx} = e^{-2\theta} \cdot \frac{-\theta + \frac{1}{\theta} + 1 + \frac{1}{\theta^2}}{\theta + \frac{1}{\theta} + 1 - \frac{1}{\theta^2}} \]

### Final Result
Thus, the final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = e^{-2\theta} \cdot \frac{-\theta + \frac{1}{\theta} + 1 + \frac{1}{\theta^2}}{\theta + \frac{1}{\theta} + 1 - \frac{1}{\theta^2}} \]