Find the critical points of the following functions.Use the Second Derivative Test to determine (if possible) whether each critical point corresponds

To find the critical point, find ${\displaystyle f_x,f_y}$ and set ${\displaystyle f_x=0,f_y=0}$
${\displaystyle f_x=t{e}^x=0\Rightarrow y=0}$ or ${\displaystyle e^x=0}$
For ${\displaystyle e^x=0}$ have no solution for ${\displaystyle x\in R}$
Therefore y=0
Set ${\displaystyle f_y=0}$
${\displaystyle f_y=e^x-e^y=0}$
Substitute ${\displaystyle y=0,f_y=e^x-e^0=0\Rightarrow e^x-1=0\Rightarrow e^x=1}$
Taking ln on both sides, ${\displaystyle \ln \left(e^x\right)=\ln \left(1\right)}$ , x=0
Thus (0,0) is the ritical point.
${\displaystyle D(x,y)=f_{\times }(x,y)f_{yy}(x,y)-{\left(f_{xy}(x,y)\right)}^2}$
1) If D(a,b)>0 and ${\displaystyle f_{\times }(a,b)>0}$ then (a,b) is local aximum of f
2) If D(a,b)>0 and ${\displaystyle f_{\times }(a,b)<0}$ then (a,b) is local aximum of f
3) If D(a,b)<0 then f(a,b) is saddle point
4) If D(a,b)=0 then this is incobclusive
Here, ${\displaystyle f_{\times }=y{e}^x,f_{yy}=-e^y}$ and ${\displaystyle f_{xy}=e^x}$
${\displaystyle D(x,y)=t{e}^x(-e^y)-{\left(e^x\right)}^2=-e{y}^{x+y}-e^{2x}}$
The critical point is(0,0)
${\displaystyle D(0,0)=-0e^0-e^{2\cdot 0}=-e^0=-1}$
Thus D(0,0)=-1<0 then f(0,0) is saddle point,
${\displaystyle f(0,0)=\left(0\right)e^0-e^0=-e^0=-1}$
Therefore -1 is saddle point.