Chapter 14 - Partial Derivatives - 14.2 Limits and Continuity - 14.2 Exercises - Page 951: 31

$D=$ {$(x,y) | x^2+y^2 \neq 1$}

As we are given that $f(x,y,z)=\dfrac{(1+x^2+y^2)}{(1-x^2-y^2)}$ The function $f(x,y,z)=\dfrac{(1+x^2+y^2)}{(1-x^2-y^2)}$ represents a rational function which is continuous on its domain $D$. Thus, $ 1-x^2-y^2 \neq 0$ or, $x^2+y^2 \neq 1$ Hence, Domain: $D=$ {$(x,y) | x^2+y^2 \neq 1$}