Chapter 14: Partial Derivatives - Section 14.3 - Partial Derivatives - Exercises 14.3 - Page 807: 10

$f_x=\frac{y^2-x^2}{(x^2+y^2)^2}$ $f_y=\frac{-2xy}{(x^2+y^2)^2}$

Take the first partial derivatives of the given function. When taking a partial derivative with respect to x, treat y as a constant, and vice versa: $f_x=\frac{(x^2+y^2)-x(2x)}{(x^2+y^2)^2}=\frac{y^2-x^2}{(x^2+y^2)^2}$ $f_y=\frac{(x^2+y^2)(0)-x(2y)}{(x^2+y^2)^2}=\frac{-2xy}{(x^2+y^2)^2}$