Largest set in which $(x, y)\mapsto \sqrt{xe^y - ye^x}$ is defined I have to find the biggest $\mathbb{R}^2$ subset in which the function 
$g(x,y) = \sqrt{xe^y - ye^x}$
is defined. In order to do it I have to study 
$xe^y - ye^x \ge 0$
but I can't find out a solution. Can someone help me? 

$$g^2(x,y)=G(x,y)=xe^y-ye^x$$ is clearly well defined when $$G(x,y)=xe^y-ye^x=0$$ which is a curve formed by the union of all the diagonal of $\mathbb R^2$ (i.e. $x=y$) and the set 0f $(x,y)$ such that $x\ne y$ satisfying the relation $$\frac xy=\frac{e^x}{e^y}$$ This second set, looking at the plot of the curve is a kind of positive branch of a hyperbola of equation $xy=a\gt0$ (but obviously not properly a hyperbola). Furthermore, it is important remark that the double point where the diagonal cuts this branch is $(x,y)=(1,1)$ which is easily calculated from
$G(x,y)=G_x(x,y)=G_y(x,y)=0$.
Now, the second quadrant is totally excluded (we excludes coordinates axis of the discussion) because
$$ x\lt0\text{ and }y\gt0\Rightarrow G(x,y)=-|x|e^{|y|}-\ |y|e^{-|x|}\lt0$$ and the fourth quadrant is totally included because of 
$$x\gt0\text{ and }y\lt0\Rightarrow G(x,y)=|x|e^{-|y|}+\ |y|e^{|x|}\gt0$$
Now we have for the third quadrant where $x$ and $y$ are negative
$$|x|\gt|y|\Rightarrow -|x|e^{-|y|}+|y|e^{-|x|}=\frac{|y|e^{|y|}-|x|e^{|x|}}{e^{|x|+|y|}}\lt0$$
$$|x|\lt |y|\Rightarrow-|x|e^{-|y|}+|y|e^{-|x|}=\frac{|y|e^{|y|}-|x|e^{|x|}}{e^{|x|+|y|}}\gt0$$
With the first quadrant, where $x$ and $y$ are positive,  the discussion is less easy and we must consider the curve $G(x,y)=0$  and its double point $(1,1)$.
From the graph of the curve $G (x, y) = 0$ seen at the beginning, it follows that for all $x\gt0$, there is $y_x\gt0$ such that $G(x, y_x)=0$. We have
$$x\le y\le y_x\text{ for }  x\le 1\\x\ge y\ge y_x\text{ for } 1\le x $$ 
In both cases one has $G(x,x)=G(x,y_x)=0$. By continuity and because for $x$ fixed the function $f_x(y)=$ $G(x,y)$ must have a minimum  in the bounded intervals $[x,y_x]$ when $x\le1$ and $[y_x,x]$ when $x\gt1$ (See explanation of this enclosed by ► and◄ below), we deduce that
$G(x,y)\lt0$ when
$x\lt y\lt y_x$ and $x\lt 1$ and when $x\gt y\gt y_x$ and $x\gt1$.
► Let $x$ be fixed, say $x=a$ positive and let the function $f_a(y)=ae^y-ye^a$. Because of $f'_a(y)=ae^y-ye^a=0$ gives $y=a-\ln(a)$ and $f''_a(x)=ae^y\gt0$ it is clear that $f_a$ has a minimum at $y=a-\ln(a)$ ◄
Finally the shape of the searched set is $A\cup B\cup C$ where
$A=$the lower half of the third quadrant.
$B=$ the fourth quadrant.
$C$ contained in the first quadrant and $C=C_1\cup C_2$ where 
$$C_1=\{(x,y):\space 0\le x\le1\text{ and }x\ge y\}\cup\{(x,y):\space 0\le x\le1\text{ and } y\ge y_x\}$$
$$C_2=\{(x,y): x\ge1\text{ and }y\le y_x\}\cup\{(x,y): x\ge1\text{ and }x\le y\}$$
NOTE.-It seems impossible to disregard the branch similar to a hyperbola branch of the curve $G (x, y) = 0$. Recall that $G (x, y_x) = 0$.

