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We are tasked with finding \(\frac{dy}{dx}\) by implicit differentiation for the equation:

\[
x e^y = x - y
\]

### Step-by-step solution:

1. **Differentiate both sides with respect to \(x\)**:

   On the left-hand side, we have the product of \(x\) and \(e^y\), so we'll apply the **product rule**:
   \[
   \frac{d}{dx}(x e^y) = \frac{d}{dx}(x) \cdot e^y + x \cdot \frac{d}{dx}(e^y)
   \]
   - The derivative of \(x\) with respect to \(x\) is 1, so the first term becomes \(e^y\).
   - The derivative of \(e^y\) with respect to \(x\) involves the chain rule:
     \[
     \frac{d}{dx}(e^y) = e^y \cdot \frac{dy}{dx}
     \]
   Thus, the derivative of the left-hand side is:
   \[
   e^y + x e^y \frac{dy}{dx}
   \]

2. **Differentiate the right-hand side**:

   The right-hand side is \(x - y\). The derivative of this with respect to \(x\) is:
   \[
   \frac{d}{dx}(x - y) = 1 - \frac{dy}{dx}
   \]

3. **Set up the equation**:

   Now, equating the derivatives of both sides:
   \[
   e^y + x e^y \frac{dy}{dx} = 1 - \frac{dy}{dx}
   \]

4. **Solve for \(\frac{dy}{dx}\)**:

   Collect all terms involving \(\frac{dy}{dx}\) on one side:
   \[
   x e^y \frac{dy}{dx} + \frac{dy}{dx} = 1 - e^y
   \]

   Factor out \(\frac{dy}{dx}\) from the left-hand side:
   \[
   \frac{dy}{dx}(x e^y + 1) = 1 - e^y
   \]

   Finally, solve for \(\frac{dy}{dx}\):
   \[
   \frac{dy}{dx} = \frac{1 - e^y}{x e^y + 1}
   \]

### Final Answer:
\[
\frac{dy}{dx} = \frac{1 - e^y}{x e^y + 1}
\]

Would you like more details, or do you have any questions about this solution?

Here are 5 related questions to deepen your understanding:
1. How does the product rule apply when differentiating two functions of \(x\)?
2. Why do we need to use the chain rule when differentiating \(e^y\)?
3. What is the geometric interpretation of implicit differentiation?
4. How would the solution change if the equation involved a different function like \(\sin(y)\) instead of \(e^y\)?
5. What happens if \(y\) is a constant? How does that simplify the differentiation?

**Tip**: Always remember to apply both the product and chain rules when differentiating composite and product terms in implicit differentiation.

Find dy/dx by implicit differentiation for the equation xe^y = x - y.

Learn how to find dy/dx by implicit differentiation for the equation xe^y = x - y. This solution applies the product rule and chain rule, making it ideal for students in grades 11-12 studying pre-calculus or calculus.

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