Answer
There is only one critical point at $\left( {0,0} \right)$.
However, by the Second Derivative Test, the critical point $\left( {0,0} \right)$ is a saddle point. Therefore, $f$ does not have a global minimum or a global maximum on the domain $D$.
This does not contradict Theorem 3 because the domain $D$ is not closed.
Work Step by Step
We have $f\left( {x,y} \right) = xy$, and the domain $D = \left\{ {\left( {x,y} \right):0 < x < 1,{\ \ }0 < y < 1} \right\}$. Notice that it is a open square. Thus, it is bounded but not closed.
The partial derivatives are
${f_x} = y$, ${\ \ \ \ }$ ${f_y} = x$
${f_{xx}} = 0$, ${\ \ }$ ${f_{yy}} = 0$, ${\ \ }$ ${f_{xy}} = 1$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = y = 0$, ${\ \ \ }$ ${f_y} = x = 0$
So, there is only one critical point at $\left( {0,0} \right)$.
We evaluate the discriminant ${f_{xx}}{f_{yy}} - {f_{xy}}^2 = - 1$. By the Second Derivative Test, the critical point $\left( {0,0} \right)$ is a saddle point.
However, the critical point $\left( {0,0} \right)$ is not defined in the domain $D = \left\{ {\left( {x,y} \right):0 < x < 1,{\ \ }0 < y < 1} \right\}$. Therefore, we conclude that $f$ does not have a global minimum or a global maximum on $D$. This does not contradict Theorem 3 because the domain $D$ is not closed.