To solve for \( A = \frac{x - y}{x + y} \) given the equation \( x^2 - 2y^2 = xy \) and the conditions \( y \neq 0 \) and \( x + y \neq 0 \), we can follow these steps:

1. Start with the given equation:
\[
x^2 - 2y^2 = xy
\]

2. Rearrange the equation to:
\[
x^2 - xy - 2y^2 = 0
\]

3. This is a quadratic equation in terms of \( x \). Solve for \( x \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -y \), and \( c = -2y^2 \):
\[
x = \frac{-(-y) \pm \sqrt{(-y)^2 - 4(1)(-2y^2)}}{2(1)}
\]
\[
x = \frac{y \pm \sqrt{y^2 + 8y^2}}{2}
\]
\[
x = \frac{y \pm \sqrt{9y^2}}{2}
\]
\[
x = \frac{y \pm 3y}{2}
\]

4. This gives us two solutions for \( x \):
\[
x = \frac{y + 3y}{2} = 2y \quad \text{and} \quad x = \frac{y - 3y}{2} = -y
\]

5. We discard the solution \( x = -y \) because it violates the condition \( x + y \neq 0 \). Therefore, we have \( x = 2y \).

6. Substitute \( x = 2y \) into \( A = \frac{x - y}{x + y} \):
\[
A = \frac{2y - y}{2y + y} = \frac{y}{3y} = \frac{1}{3}
\]

Thus, the value of \( A \) is:
\[
A = \frac{1}{3}
\]