We are given the following equations:

$z = \frac{y}{x}, \quad x = e^t, \quad y = 1 - e^{2t}$

We need to find $\frac{dz}{dt}$.

Step 1: Express $z$ in terms of $t$

First, substitute the expressions for $x$ and $y$ into the equation for $z$:

$z = \frac{y}{x} = \frac{1 - e^{2t}}{e^t}$

We can simplify this expression:

$z = \frac{1}{e^t} - \frac{e^{2t}}{e^t} = e^{-t} - e^t$

Step 2: Differentiate $z$ with respect to $t$

Now, differentiate $z = e^{-t} - e^t$ with respect to $t$:

$\frac{dz}{dt} = \frac{d}{dt}\left( e^{-t} \right) - \frac{d}{dt}\left( e^t \right)$

Using the chain rule:

$\frac{dz}{dt} = -e^{-t} - e^t$

Thus, the derivative of $z$ with respect to $t$ is:

$\frac{dz}{dt} = -e^{-t} - e^t$

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Would you like more details or have any questions on this process?

Here are some related questions for further exploration:

1. How would the result change if $x = e^{2t}$ instead of $x = e^t$?
2. How do the rules of differentiation apply when combining exponential functions?
3. What is the relationship between the derivatives of $x$ and $y$ in this problem?
4. How would you apply the chain rule in other types of functions like logarithmic or trigonometric?
5. Can you differentiate implicitly if $z = \frac{y}{x}$ and the relationships between $x$ and $y$ are more complicated?

Tip: When differentiating exponential functions, remember that $\frac{d}{dt} e^{at} = ae^{at}$, which is a helpful shortcut!