Answer
There is one critical point at $\left( {\frac{1}{2},\frac{1}{2}} \right)$. It is a saddle point.
Work Step by Step
We have $f\left( {x,y} \right) = {{\rm{e}}^{x + y}} - x{{\rm{e}}^{2y}}$.
The partial derivatives are
${f_x} = {{\rm{e}}^{x + y}} - {{\rm{e}}^{2y}}$, ${\ \ \ }$ ${f_y} = {{\rm{e}}^{x + y}} - 2x{{\rm{e}}^{2y}}$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = {{\rm{e}}^{x + y}} - {{\rm{e}}^{2y}} = 0$
${{\rm{e}}^{x + y}} = {{\rm{e}}^{2y}}$
${f_y} = {{\rm{e}}^{x + y}} - 2x{{\rm{e}}^{2y}} = 0$
${{\rm{e}}^{x + y}} = 2x{{\rm{e}}^{2y}}$
From these two equations we obtain
${{\rm{e}}^{2y}} = 2x{{\rm{e}}^{2y}}$
${{\rm{e}}^{2y}}\left( {1 - 2x} \right) = 0$
Since ${{\rm{e}}^{2y}} \ne 0$, we have $1 - 2x = 0$. So, $x = \frac{1}{2}$.
Substituting $x = \frac{1}{2}$ in ${{\rm{e}}^{x + y}} = {{\rm{e}}^{2y}}$ gives
${{\rm{e}}^{\frac{1}{2} + y}} = {{\rm{e}}^{2y}}$
${{\rm{e}}^y}\left( {{{\rm{e}}^{1/2}} - {{\rm{e}}^y}} \right) = 0$
Since ${{\rm{e}}^y} \ne 0$, we have ${{\rm{e}}^{1/2}} - {{\rm{e}}^y} = 0$. So, $y = \frac{1}{2}$.
So, the critical point is $\left( {\frac{1}{2},\frac{1}{2}} \right)$.
The Second Derivative Test:
We evaluate the second partial derivatives:
${f_{xx}} = {{\rm{e}}^{x + y}}$, ${\ \ \ }$ ${f_{yy}} = {{\rm{e}}^{x + y}} - 4x{{\rm{e}}^{2y}}$, ${\ \ \ }$ ${f_{xy}} = {{\rm{e}}^{x + y}} - 2{{\rm{e}}^{2y}}$
At the critical point $\left( {\frac{1}{2},\frac{1}{2}} \right)$ we get
${f_{xx}} = {\rm{e}}$, ${\ \ \ }$ ${f_{yy}} = - {\rm{e}}$, ${\ \ \ }$ ${f_{xy}} = - {\rm{e}}$
$D = {f_{xx}}{f_{yy}} - {f_{xy}}^2 = - 2{{\rm{e}}^2}$
Since $D<0$, $f$ has a saddle point at $\left( {\frac{1}{2},\frac{1}{2}} \right)$.
