Answer
$h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$;
set of points: {$(x,y) | 1\gt xy$}
Work Step by Step
We are given: $h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$
The function is continuous everywhere and the $ln$ term is only defined for positive 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: {$(x,y) | 1\gt xy$}
