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To solve for \( y' \) at the point where \( x = 0 \) for the given equation:

\[
xy + 8e^y = 8e
\]

we'll go through the following steps:

### Step 1: Differentiate both sides with respect to \( x \)
Since the equation involves both \( x \) and \( y \), we'll apply implicit differentiation. Differentiate each term:

\[
\frac{d}{dx}(xy) + \frac{d}{dx}(8e^y) = \frac{d}{dx}(8e)
\]

1. For \( \frac{d}{dx}(xy) \): Use the product rule:
   \[
   \frac{d}{dx}(xy) = y + x \frac{dy}{dx} = y + x y'
   \]

2. For \( \frac{d}{dx}(8e^y) \): Use the chain rule:
   \[
   \frac{d}{dx}(8e^y) = 8e^y \cdot y'
   \]

3. For \( \frac{d}{dx}(8e) \): This is a constant, so its derivative is 0.

Putting it all together:
\[
y + x y' + 8e^y y' = 0
\]

### Step 2: Solve for \( y' \)
Combine terms involving \( y' \):
\[
x y' + 8e^y y' = -y
\]

Factor out \( y' \):
\[
y'(x + 8e^y) = -y
\]

Thus,
\[
y' = \frac{-y}{x + 8e^y}
\]

### Step 3: Substitute \( x = 0 \)
We are asked to find \( y' \) at \( x = 0 \). Substitute \( x = 0 \) into the equation:

\[
y' = \frac{-y}{0 + 8e^y} = \frac{-y}{8e^y}
\]

### Step 4: Find the value of \( y \) when \( x = 0 \)
Use the original equation to find \( y \) when \( x = 0 \):

\[
0 \cdot y + 8e^y = 8e
\]

This simplifies to:
\[
8e^y = 8e
\]

Divide both sides by 8:
\[
e^y = e
\]

Taking the natural logarithm of both sides:
\[
y = 1
\]

### Step 5: Substitute \( y = 1 \) into the expression for \( y' \)
Now that we know \( y = 1 \) when \( x = 0 \), substitute \( y = 1 \) into the expression for \( y' \):

\[
y' = \frac{-1}{8e^1} = \frac{-1}{8e}
\]

### Final Answer:
\[
y' = -\frac{1}{8e} \quad \text{at} \quad x = 0
\]

Would you like further details or have any questions?

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Here are some related questions to explore further:
1. How does implicit differentiation work in general cases?
2. What are the rules of differentiation for exponential functions?
3. Can we solve this problem using a different approach, such as solving for \( y \) in terms of \( x \)?
4. What happens to \( y' \) as \( x \) approaches a different value, like \( x = 1 \)?
5. How would the answer change if the equation were \( xy + 8e^y = 10e \)?

**Tip:** When differentiating implicitly, keep track of each term carefully, especially when using product and chain rules.

If xy + 8e^y = 8e, find the value of y' at the point where x = 0.

Solve for y' using implicit differentiation in the equation xy + 8e^y = 8e at x = 0. This solution includes the product and chain rules, along with key exponential function concepts. Ideal for high school calculus students.

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