1. **State the problem:** Find the second order partial derivatives of the function $$f(x,y) = \frac{x - y}{x^2 - y^2}$$. 2. **Simplify the function:** Note that the denominator can be factored as $$x^2 - y^2 = (x - y)(x + y)$$. So, $$f(x,y) = \frac{x - y}{(x - y)(x + y)} = \frac{1}{x + y}, \quad x \neq y$$ 3. **Find the first order partial derivatives:** - Partial derivative with respect to $$x$$: $$f_x = \frac{\partial}{\partial x} \left( \frac{1}{x + y} \right) = -\frac{1}{(x + y)^2}$$ - Partial derivative with respect to $$y$$: $$f_y = \frac{\partial}{\partial y} \left( \frac{1}{x + y} \right) = -\frac{1}{(x + y)^2}$$ 4. **Find the second order partial derivatives:** - Second partial derivative with respect to $$x$$: $$f_{xx} = \frac{\partial}{\partial x} \left(-\frac{1}{(x + y)^2} \right) = 2 \frac{1}{(x + y)^3}$$ - Second partial derivative with respect to $$y$$: $$f_{yy} = \frac{\partial}{\partial y} \left(-\frac{1}{(x + y)^2} \right) = 2 \frac{1}{(x + y)^3}$$ - Mixed partial derivatives: $$f_{xy} = \frac{\partial}{\partial y} \left(-\frac{1}{(x + y)^2} \right) = 2 \frac{1}{(x + y)^3}$$ $$f_{yx} = \frac{\partial}{\partial x} \left(-\frac{1}{(x + y)^2} \right) = 2 \frac{1}{(x + y)^3}$$ 5. **Summary:** $$f_{xx} = f_{yy} = f_{xy} = f_{yx} = \frac{2}{(x + y)^3}$$ These are the second order partial derivatives of the given function.