To solve this, we calculate the partial derivative of the given function $f(x, y) = xy + e^{x-y}$ with respect to $y$, and then evaluate it at $(x, y) = (0, 1)$.

Step 1: Partial derivative with respect to $y$

The function $f(x, y)$ is given as: $f(x, y) = xy + e^{x-y}$

To compute $\frac{\partial f}{\partial y}$, differentiate $f(x, y)$ with respect to $y$, treating $x$ as a constant.

1. Differentiate $xy$ with respect to $y$:
   $\frac{\partial}{\partial y}(xy) = x$

2. Differentiate $e^{x-y}$ with respect to $y$:
   Using the chain rule,
   $\frac{\partial}{\partial y}(e^{x-y}) = e^{x-y} \cdot \frac{\partial}{\partial y}(-y) = -e^{x-y}$

Thus,
$\frac{\partial f}{\partial y} = x - e^{x-y}$

Step 2: Evaluate at $(x, y) = (0, 1)$

Substitute $x = 0$ and $y = 1$ into the partial derivative: $\frac{\partial f}{\partial y}(0, 1) = 0 - e^{0-1}$

Simplify: $\frac{\partial f}{\partial y}(0, 1) = -e^{-1}$

Final Answer:

The value of $\frac{\partial f}{\partial y}(0, 1)$ is $-e^{-1}$.

Thus, the correct option is: C. $-e^{-1}$

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Would you like further clarification or details about this calculation?

Related Questions:

1. What are the steps to compute partial derivatives in multivariable functions?
2. How is the chain rule applied in partial differentiation?
3. Can you explain the significance of evaluating derivatives at specific points?
4. What is the interpretation of $\frac{\partial f}{\partial y}$ in the context of this problem?
5. How do exponential functions behave in differentiation?

Tip:

When computing partial derivatives, carefully track which variables are held constant and apply the chain rule consistently!