Find all pairs of relatively prime numbers ($x, y$) such that $x^2(x + y)$ is divisible by $y^2(y - x)^2$.
.
To find all pairs of relatively prime numbers \((x, y)\) such that \(x^2(x + y)\) is divisible by \(y^2(y - x)^2\), we proceed as follows:

1. **Divisibility Condition**:
   Given that \(x^2(x + y)\) is divisible by \(y^2(y - x)^2\), we can write:
   \[
   y^2(y - x)^2 \mid x^2(x + y)
   \]

2. **Simplifying the Divisibility**:
   Since \(y^2(y - x)^2\) divides \(x^2(x + y)\), it implies:
   \[
   y^2(y - x)^2 \mid x^2(x + y)
   \]
   This can be expanded and simplified to:
   \[
   y^2(y - x)^2 \mid x^2(x + y) \implies y^2(y^2 - 2xy + x^2) \mid x^2(x + y)
   \]

3. **Analyzing the Divisibility**:
   Since \(y^2(y^2 - 2xy + x^2)\) must divide \(x^2(x + y)\), we can consider the prime factors of \(y\). If \(y\) divides \(x^2(x + y)\), then:
   \[
   y \mid x^2(x + y)
   \]
   Since \(x\) and \(y\) are relatively prime, \(y\) must divide \(x + y\). Therefore, \(y\) must divide \(x\).

4. **Relatively Prime Condition**:
   Since \(x\) and \(y\) are relatively prime, the only way \(y\) can divide \(x\) is if \(y = 1\). Thus, we have:
   \[
   y = 1
   \]

5. **Substituting \(y = 1\)**:
   Substituting \(y = 1\) into the original condition, we get:
   \[
   x^2(x + 1) \text{ must be divisible by } 1^2(1 - x)^2 = (1 - x)^2
   \]
   Simplifying, we have:
   \[
   (1 - x)^2 \mid x^2(x + 1)
   \]

6. **Checking the Divisibility**:
   We need to check when \((1 - x)^2\) divides \(x^2(x + 1)\). This implies:
   \[
   (1 - x)^2 \mid x^2(x + 1)
   \]
   Since \((1 - x)\) and \(x\) are relatively prime, we need to check the possible values of \(x\):
   \[
   1 - x \mid x + 1
   \]
   This implies:
   \[
   1 - x \mid 2
   \]

7. **Solving the Divisibility**:
   The possible values for \(1 - x\) that divide 2 are \(\pm 1, \pm 2\). Thus, we have:
   \[
   1 - x = 1 \implies x = 0 \quad (\text{not valid since } x \text{ and } y \text{ are relatively prime})
   \]
   \[
   1 - x = -1 \implies x = 2
   \]
   \[
   1 - x = 2 \implies x = -1 \quad (\text{not valid since } x \text{ and } y \text{ are relatively prime})
   \]
   \[
   1 - x = -2 \implies x = 3
   \]

8. **Valid Solutions**:
   Therefore, the valid solutions are:
   \[
   (x, y) = (2, 1) \quad \text{and} \quad (x, y) = (3, 1)
   \]

The final answer is \( \boxed{ (x, y) = (2, 1) } \) and \((x, y) = (3, 1)\).