To eliminate the parameter $t$ from the equations $x = e^{-3t}$ and $y = e^{3t}$ and find a Cartesian equation, we can follow these steps:

1. Start with the given equations: $x = e^{-3t} \quad \text{and} \quad y = e^{3t}$

2. Take the natural logarithm of both sides of each equation to solve for $t$: $\ln(x) = -3t \quad \Rightarrow \quad t = -\frac{\ln(x)}{3}$ and $\ln(y) = 3t \quad \Rightarrow \quad t = \frac{\ln(y)}{3}$

3. Since both expressions equal $t$, set them equal to each other: $-\frac{\ln(x)}{3} = \frac{\ln(y)}{3}$

4. Multiply both sides by 3 to eliminate the denominator: $-\ln(x) = \ln(y)$

5. Rewrite the equation by taking exponentials: $\ln\left(\frac{1}{x}\right) = \ln(y) \quad \Rightarrow \quad \frac{1}{x} = y$

Thus, the Cartesian equation of the curve is: $y = \frac{1}{x}$

This equation describes the relationship between $x$ and $y$ on the curve after eliminating the parameter $t$.

Would you like further details or have any questions?

Here are some related questions to explore:

1. How would the solution change if $x = e^{2t}$ and $y = e^{-2t}$?
2. What is the domain of the function $y = \frac{1}{x}$?
3. How would you find the tangent line to this curve at a specific point?
4. How does changing the signs of the exponents in $x$ and $y$ affect the Cartesian equation?
5. Can you graph the equation $y = \frac{1}{x}$ and describe its behavior?

Tip: Remember that logarithmic operations are helpful in converting exponential expressions, especially when eliminating parameters from parametric equations.