131. If \( f(x, y)=\frac{x^{2}-2 x y+y^{2}}{x^{2}-y^{2}} \), find \( f(-x,-y) \)

To find \( f(-x, -y) \), we need to substitute \(-x\) for \(x\) and \(-y\) for \(y\) in the given function \( f(x, y) = \frac{x^{2}-2xy+y^{2}}{x^{2}-y^{2}} \). First, let's compute the numerator with the substitutions: \[ f(-x, -y) = \frac{(-x)^{2} - 2(-x)(-y) + (-y)^{2}}{(-x)^{2} - (-y)^{2}} \] Calculating the components: 1. **Numerator**: \[ (-x)^{2} = x^{2}, \quad -2(-x)(-y) = -2xy, \quad (-y)^{2} = y^{2} \] Therefore, the numerator becomes: \[ x^{2} - 2xy + y^{2} \] 2. **Denominator**: \[ (-x)^{2} = x^{2}, \quad (-y)^{2} = y^{2} \] Thus, the denominator is: \[ x^{2} - y^{2} \] Now, we can put it all together: \[ f(-x, -y) = \frac{x^{2} - 2xy + y^{2}}{x^{2} - y^{2}} \] To relate this back to the original function, notice that the numerator can be rewritten: \[ x^{2} - 2xy + y^{2} = (x - y)^{2} \] Thus, we have: \[ f(-x, -y) = \frac{(x - y)^{2}}{x^{2} - y^{2}} \] Since \(x^{2} - y^{2} = (x - y)(x + y)\), we can simplify further: \[ f(-x, -y) = \frac{(x - y)^{2}}{(x - y)(x + y)} = \frac{x - y}{x + y} \text{ when } x \neq y \] So the final answer is: \[ f(-x, -y) = \frac{x - y}{x + y} \]