`{:((a^(2))/(x) - (b^(2))/(y) = 0),((a^(2)b)/(x)+(b^(2)a)/(y) = "a + b, where x, y"ne 0.):}`

Answer

To solve the given equations:

1. **Equations**:
\[ \frac{a^2}{x} - \frac{b^2}{y} = 0 \]
\[ \frac{a^2 b}{x} + \frac{b^2 a}{y} = a + b \]

2. **Rearranging the first equation**:
From the first equation, we can rearrange it to express one variable in terms of the other:
\[ \frac{a^2}{x} = \frac{b^2}{y} \]
Cross-multiplying gives:
\[ a^2 y = b^2 x \]
Thus,
\[ \frac{1}{x} = \frac{b^2}{a^2 y} \]

3. **Substituting**:
Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). Hence, we can rewrite the first equation as:
\[ a^2 u = b^2 v \quad \text{(1)} \]

4. **Rearranging the second equation**:
Substitute \( u \) and \( v \) into the second equation:
\[ \frac{a^2 b}{x} + \frac{b^2 a}{y} = a + b \]
This becomes:
\[ a^2 b u + b^2 a v = a + b \]

5. **Substituting for \( u \)**:
From equation (1), we have \( u = \frac{b^2 v}{a^2} \). Substitute this into the second equation:
\[ a^2 b \left(\frac{b^2 v}{a^2}\right) + b^2 a v = a + b \]
Simplifying gives:
\[ b^3 v + b^2 a v = a + b \]
Factoring out \( v \):
\[ v(b^3 + b^2 a) = a + b \]

6. **Solving for \( v \)**:
Thus,
\[ v = \frac{a + b}{b^2 (b + a)} \]
The \( a + b \) terms cancel out:
\[ v = \frac{1}{b^2} \]

7. **Finding \( u \)**:
Substitute \( v \) back into the expression for \( u \):
\[ u = \frac{b^2 v}{a^2} = \frac{b^2 \cdot \frac{1}{b^2}}{a^2} = \frac{1}{a^2} \]

8. **Finding \( x \) and \( y \)**:
Since \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \):
\[ \frac{1}{x} = \frac{1}{a^2} \implies x = a^2 \]
\[ \frac{1}{y} = \frac{1}{b^2} \implies y = b^2 \]

9. **Final Answers**:
Therefore, the solutions are:
\[ x = a^2, \quad y = b^2 \]