1. **State the problem:** Find the second order partial derivatives of the function $$f(x,y) = \frac{x - y}{x^2 - y^2}$$ with respect to $x$ and $y$. 2. **Simplify the function:** Note that the denominator can be factored as a difference of squares: $$x^2 - y^2 = (x - y)(x + y)$$ 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 derivative: $$f_{xy} = \frac{\partial}{\partial y} \left( -\frac{1}{(x + y)^2} \right) = 2 \frac{1}{(x + y)^3}$$ 5. **Final answers:** $$f_{xx} = \frac{2}{(x + y)^3}, \quad f_{yy} = \frac{2}{(x + y)^3}, \quad f_{xy} = \frac{2}{(x + y)^3}$$