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The problem in the image asks you to compute \(\frac{dz}{dt}\), where \( z = f(x, y) = xye^{xy} \), with \( x(t) = t \) and \( y(t) = \frac{1}{t} \).

### Step-by-Step Solution:

We are tasked with finding \(\frac{dz}{dt}\), and to do this, we will use the chain rule since \( z = f(x, y) \) is a function of \( x \) and \( y \), which in turn are functions of \( t \).

1. **Restate the function:**
   \[
   z = xye^{xy}
   \]
   where \( x = t \) and \( y = \frac{1}{t} \).

2. **Apply the chain rule:**
   \[
   \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}
   \]

3. **Compute partial derivatives:**

   - For \(\frac{\partial z}{\partial x}\), treat \( y \) as a constant:
     \[
     \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \left( xye^{xy} \right) = ye^{xy} + xy^2e^{xy}
     \]
   
   - For \(\frac{\partial z}{\partial y}\), treat \( x \) as a constant:
     \[
     \frac{\partial z}{\partial y} = \frac{\partial}{\partial y} \left( xye^{xy} \right) = xe^{xy} + x^2ye^{xy}
     \]

4. **Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):**
   - Since \( x(t) = t \), we have \(\frac{dx}{dt} = 1\).
   - Since \( y(t) = \frac{1}{t} \), we have:
     \[
     \frac{dy}{dt} = -\frac{1}{t^2}
     \]

5. **Substitute \( x = t \) and \( y = \frac{1}{t} \) into the partial derivatives:**

   - For \(\frac{\partial z}{\partial x}\):
     \[
     \frac{\partial z}{\partial x} = \frac{1}{t} e^1 + t \left( \frac{1}{t^2} \right) e^1 = \frac{1}{t} e + \frac{e}{t}
     \]
     This simplifies to:
     \[
     \frac{\partial z}{\partial x} = \frac{2e}{t}
     \]

   - For \(\frac{\partial z}{\partial y}\):
     \[
     \frac{\partial z}{\partial y} = t e^1 + t^2 \left( \frac{1}{t} \right) e^1 = te + te
     \]
     This simplifies to:
     \[
     \frac{\partial z}{\partial y} = 2te
     \]

6. **Substitute into the chain rule expression:**
   \[
   \frac{dz}{dt} = \left( \frac{2e}{t} \right)(1) + (2te) \left( -\frac{1}{t^2} \right)
   \]
   Simplify:
   \[
   \frac{dz}{dt} = \frac{2e}{t} - \frac{2e}{t}
   \]
   Thus:
   \[
   \frac{dz}{dt} = 0
   \]

### Final Answer:
\[
\frac{dz}{dt} = 0
\]

Would you like more details or further explanations?

### Related Questions:
1. How does the chain rule apply to functions of multiple variables?
2. What is the significance of partial derivatives in finding the total derivative?
3. Can you explain why \( \frac{dz}{dt} = 0 \) in this problem?
4. How would the solution change if \( y(t) \) was a different function of \( t \)?
5. How do exponential functions affect differentiation?

### Tip:
Always keep track of how each variable depends on the independent variable when using the chain rule with functions of multiple variables!

Suppose z = f(x, y) = xye^{xy} and x(t) = t and y(t) = 1/t. Compute dz/dt.

Learn how to compute dz/dt using the chain rule and partial derivatives for the function z = xye^{xy}, where x(t) = t and y(t) = 1/t. Step-by-step explanation of the solution, covering key calculus concepts and formulas. Suitable for Calculus students at the undergraduate level.

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