Chapter 14 - Partial Derivatives - 14.2 Exercises - Page 924: 34

{$(x,y) | x+y \neq 0$}

As we are given that $f(x,y)=(x+y)^{-2}$ The function $f(x,y)=(x+y)^{-2}$ can be re-written as $f(x,y)=\dfrac{1}{(x+y)^{2}}$ which will be undefined only for $(x+y)^2=0$ and exists only for positive numbers. Thus, $ x+y \neq 0$ Hence, {$(x,y) | x+y \neq 0$}