## Calculus 8th Edition

$h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$; set of points: {$(x,y) | 1\gt xy$}
As per the given values; $h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$ The function is continuous everywhere and the term $\dfrac{(1-xy)}{(1+x^2y^2)}$ is greater than $0$ such as $t \gt 0$ these shows the number of values of $t$ for which $\ln t$ exist for such values. Thus, $1-xy\gt 0$ or, $1\gt xy$ Hence, $h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$; set of points: {$(x,y) | 1\gt xy$}