To solve this problem, we are tasked to prove:

$\frac{d^2y}{dx^2} = 2 \left( \frac{y}{x} \right)^3$

given the equation $x + y = xy$. Here's the step-by-step solution:

Step 1: Rewrite the equation

The given equation is:

$x + y = xy$

Rearrange it to isolate $y$:

$xy - y = x$

$y(x - 1) = x$

$y = \frac{x}{x - 1}$

Step 2: First derivative $\frac{dy}{dx}$

Using the quotient rule for differentiation:

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

Simplify:

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

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

Step 3: Second derivative $\frac{d^2y}{dx^2}$

Differentiate $\frac{dy}{dx}$ again. Using the chain rule:

$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{-1}{(x-1)^2} \right)$

Differentiate:

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

Using the power rule:

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

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

$\frac{d^2y}{dx^2} = 2(x-1)^{-3}$

Step 4: Rewrite in terms of $y$

From Step 1, we know that:

$y = \frac{x}{x-1}$

Rearranging gives:

$x - 1 = \frac{x}{y}$

Substitute $x-1 = \frac{x}{y}$ into $\frac{d^2y}{dx^2}$:

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

$\frac{d^2y}{dx^2} = 2 \left( \frac{1}{\left(\frac{x}{y}\right)^3} \right)$

Simplify:

$\frac{d^2y}{dx^2} = 2 \left( \frac{y^3}{x^3} \right)$

$\frac{d^2y}{dx^2} = 2 \left( \frac{y}{x} \right)^3$

Final Answer:

$\frac{d^2y}{dx^2} = 2 \left( \frac{y}{x} \right)^3$

Would you like a detailed explanation of any specific step or additional examples?

Here are 5 related questions for further exploration:

1. How does the quotient rule work in calculus, and why is it used here?
2. What are the conditions under which $y = \frac{x}{x-1}$ is valid?
3. Can we solve this problem using implicit differentiation instead of explicit $y$?
4. How does substituting back into terms of $y$ simplify the result?
5. Why does the second derivative $\frac{d^2y}{dx^2}$ describe the curvature of $y$?

Tip: Always simplify the given equation first to make derivatives easier to compute.