Answer
The critical point and its nature:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{}\\
{{\rm{Point}}}&{{\rm{Type}}}\\
{\left( {0,0} \right)}&{{\rm{saddle{\ } point}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = {{\rm{e}}^x} - x{{\rm{e}}^y}$.
The partial derivatives are
${f_x} = {{\rm{e}}^x} - {{\rm{e}}^y}$, ${\ \ \ }$ ${f_y} = - x{{\rm{e}}^y}$
${f_{xx}} = {{\rm{e}}^x}$, ${\ \ }$ ${f_{yy}} = - x{{\rm{e}}^y}$, ${\ \ }$ ${f_{xy}} = - {{\rm{e}}^y}$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = {{\rm{e}}^x} - {{\rm{e}}^y} = 0$, ${\ \ \ }$ ${f_y} = - x{{\rm{e}}^y} = 0$
The solution of the first equation is $y=x$.
Substituting $y=x$ in the second equation gives $x=0$.
So, there is only one critical point: $\left( {0,0} \right)$.
We use the Second Derivative Test to determine the nature of the critical point. The results are listed in the following table:
$\begin{array}{*{20}{c}}
{{\bf{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\bf{Discriminant}}}&{}\\
{{\bf{Point}}}&{{{\rm{e}}^x}}&{ - x{{\rm{e}}^y}}&{ - {{\rm{e}}^y}}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\rm{Type}}}\\
{\left( {0,0} \right)}&1&0&{ - 1}&{ - 1}&{{\rm{saddle{\ } point}}}
\end{array}$
