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The given problem is:

设 \( y = y(x) \) 是由方程 \( xy - \ln y = 1 \) 所确定的隐函数，求 \( y'(0) \);

To solve this, we need to differentiate the given implicit function \( xy - \ln y = 1 \) with respect to \( x \).

1. Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx} (xy) - \frac{d}{dx} (\ln y) = \frac{d}{dx} (1) \]

2. Apply the product rule and the chain rule:
\[ y + x \frac{dy}{dx} - \frac{1}{y} \frac{dy}{dx} = 0 \]

3. Combine terms involving \( \frac{dy}{dx} \):
\[ y + \left( x - \frac{1}{y} \right) \frac{dy}{dx} = 0 \]

4. Solve for \( \frac{dy}{dx} \):
\[ \left( x - \frac{1}{y} \right) \frac{dy}{dx} = -y \]
\[ \frac{dy}{dx} = \frac{-y}{x - \frac{1}{y}} \]
\[ \frac{dy}{dx} = \frac{-y^2}{xy - 1} \]

We are asked to find \( y'(0) \). To do this, we need the value of \( y \) when \( x = 0 \).

From the original equation:
\[ x \cdot y - \ln y = 1 \]
Substitute \( x = 0 \):
\[ 0 \cdot y - \ln y = 1 \]
\[ - \ln y = 1 \]
\[ \ln y = -1 \]
\[ y = e^{-1} = \frac{1}{e} \]

Now, substitute \( x = 0 \) and \( y = \frac{1}{e} \) into the expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{-(\frac{1}{e})^2}{0 \cdot \frac{1}{e} - 1} \]
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{-\frac{1}{e^2}}{-1} \]
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{1}{e^2} \]

Thus, \( y'(0) = \frac{1}{e^2} \).

Do you want more details or have any questions? Here are 5 questions you might consider next:

1. How do you apply the chain rule in implicit differentiation?
2. Can you explain the product rule used in differentiation?
3. What are some common methods for solving implicit differentiation problems?
4. How do you find the second derivative of an implicitly defined function?
5. What are the applications of implicit differentiation in real-world problems?

**Tip:** When differentiating implicitly, always remember to apply the chain rule to terms involving \( y \).

Learn how to find the derivative y'(0) for the implicit function xy - ln y = 1 using implicit differentiation. Detailed step-by-step solution involving differentiation rules and exponential functions. Suitable for advanced high school students.

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