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We are given the equation:

\[
x + y = xy
\]

We need to show that:

\[
\frac{d^2 y}{dx^2} = \frac{2y^3}{x^3}
\]

### Step 1: Rearrange the original equation

Starting from \(x + y = xy\), we can solve for \(y\) in terms of \(x\):

\[
x + y = xy \quad \Rightarrow \quad x = xy - y
\]
\[
x = y(x - 1)
\]
\[
y = \frac{x}{x - 1}
\]

### Step 2: Differentiate \(y\) with respect to \(x\)

Now, differentiate \(y = \frac{x}{x-1}\) using the quotient rule:

\[
\frac{dy}{dx} = \frac{(x - 1)(1) - x(1)}{(x - 1)^2} = \frac{x - 1 - x}{(x - 1)^2} = \frac{-1}{(x - 1)^2}
\]

### Step 3: Differentiate again to get \( \frac{d^2y}{dx^2} \)

Differentiate \(\frac{dy}{dx} = \frac{-1}{(x - 1)^2}\) again using the chain rule:

\[
\frac{d^2 y}{dx^2} = \frac{d}{dx}\left( \frac{-1}{(x - 1)^2} \right)
\]
\[
\frac{d^2 y}{dx^2} = 2 \cdot \frac{1}{(x - 1)^3}
\]

### Step 4: Express the second derivative in terms of \(y\)

Since \(y = \frac{x}{x - 1}\), we can express \(x - 1\) as:

\[
x - 1 = \frac{x}{y} \quad \Rightarrow \quad \frac{d^2 y}{dx^2} = \frac{2}{\left(\frac{x}{y}\right)^3} = \frac{2y^3}{x^3}
\]

Thus, we have shown that:

\[
\frac{d^2 y}{dx^2} = \frac{2y^3}{x^3}
\]

Let me know if you want further clarifications or details!

### Related questions:
1. How do you apply the quotient rule for differentiation?
2. Why is the chain rule important in higher-order derivatives?
3. What other ways can this equation \(x + y = xy\) be rearranged?
4. How do you handle implicit differentiation for equations like \(x + y = xy\)?
5. How can we generalize this result for different equations?

#### Tip:
Always express quantities in simpler terms (like expressing \(x - 1\) as \(\frac{x}{y}\)) before performing complex operations like differentiation.

If x + y = xy, show that d^2y/dx^2 = 2y^3/x^3.

Learn how to solve the differential calculus problem where x + y = xy, and prove that the second derivative d^2y/dx^2 = 2y^3/x^3. This step-by-step solution involves algebraic manipulation, the quotient rule, and chain rule.

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