Consider the function $h(x,y) = (x^2+y^2)^{xy}; (x,y)\in {\Re}^{2} $
1- Determine $D_h$ the domain of definition of $h$.
2- Plot in ${\Re}^{2}$ the level curve $\Gamma= \{(x,y) \ | \ h(x,y)=1 \}$
A1 - When it comes to finding the domain of definition, it is obvious that the function is not defined for $\{(x, y)\ | \ x^2+y^2 = 0 \ \land \ xy = 0 \}$
So $D_h= {\Re}^{2} \backslash(0,0)$
A2 - For every element of $\Gamma$, it is enough that $x = 0$ or $y = 0$ or $x^2+y^2 =1$
My question is: shall we include $(0,0)$ in $\Gamma$ and write $$\Gamma = \{(x,y) | x=0 \} \cup \{ (x,y)| y=0 \} \cup \{(x,y) | x^2+y^2=1 \}$$
Or just exclude it since it is out of the domain of definition of our function, and write: $$\Gamma = \{(x,y) | x=0 \} \cup \{ (x,y)| y=0 \} \cup \{(x,y) | x^2+y^2=1 \} \backslash(0,0)$$
This depends upon how you define $a^b$, with $a,b\in\Bbb R$. If you define it as $e^{b\log a}$, then it is defined if and only if $a>0$ (with no restriction on $b$). Therefore, $h$ is defined if and only if $(x,y)\ne(0,0)$. In other words, the domain of $h$ is indeed $\Bbb R^2\setminus\{(0,0)\}$.
But then $\Gamma$ can consist only of points of $\Bbb R^2\setminus\{(0,0)\}$. And\begin{align}h(x,y)=1&\iff(x^2+y^2)^{xy}=1\\&\iff e^{xy\log(x^2+y^2)}=1\\&\iff xy\log(x^2+y^2)=0\\&\iff x=0\vee y=0.\end{align}So,$$\Gamma=\{(x,y)\in\Bbb R^2\setminus\{(0,0)\}\mid x=0\vee y=0\}.$$