Answer
The critical point and its nature:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{}\\
{{\rm{Point}}}&{{\rm{Type}}}\\
{\left( {0,2} \right)}&{{\rm{saddle{\ } point}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = {{\rm{e}}^{{x^2} - {y^2} + 4y}}$.
The partial derivatives are
${f_x} = 2x{{\rm{e}}^{{x^2} - {y^2} + 4y}}$, ${\ \ }$ ${f_y} = \left( { - 2y + 4} \right){{\rm{e}}^{{x^2} - {y^2} + 4y}}$
${f_{xx}} = 2{{\rm{e}}^{{x^2} - {y^2} + 4y}} + 4{x^2}{{\rm{e}}^{{x^2} - {y^2} + 4y}}$
$ = \left( {2 + 4{x^2}} \right){{\rm{e}}^{{x^2} - {y^2} + 4y}}$
${f_{yy}} = - 2{{\rm{e}}^{{x^2} - {y^2} + 4y}} + {\left( { - 2y + 4} \right)^2}{{\rm{e}}^{{x^2} - {y^2} + 4y}}$
$ = \left( { - 2 + {{\left( { - 2y + 4} \right)}^2}} \right){{\rm{e}}^{{x^2} - {y^2} + 4y}}$
${f_{xy}} = 2x\left( { - 2y + 4} \right){{\rm{e}}^{{x^2} - {y^2} + 4y}}$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 2x{{\rm{e}}^{{x^2} - {y^2} + 4y}} = 0$, ${\ \ }$ ${f_y} = \left( { - 2y + 4} \right){{\rm{e}}^{{x^2} - {y^2} + 4y}} = 0$
Since ${{\rm{e}}^{{x^2} - {y^2} + 4y}} \ne 0$, the solutions are $x=0$ and $y=2$.
So, there is only one critical point: $\left( {0,2} \right)$.
Recall from previous results:
${f_{xx}} = \left( {2 + 4{x^2}} \right){{\rm{e}}^{{x^2} - {y^2} + 4y}}$
${f_{yy}} = \left( { - 2 + {{\left( { - 2y + 4} \right)}^2}} \right){{\rm{e}}^{{x^2} - {y^2} + 4y}}$
${f_{xy}} = 2x\left( { - 2y + 4} \right){{\rm{e}}^{{x^2} - {y^2} + 4y}}$
We evaluate ${f_{xx}}$, ${f_{yy}}$ and ${f_{xy}}$ at the critical point and use the Second Derivative Test to determine the nature of the critical point. The results are listed in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{}&{}&{}&{{\rm{Discriminant}}}&{}\\
{{\rm{Point}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\rm{Type}}}\\
{\left( {0,2} \right)}&{2{{\rm{e}}^4}}&{ - 2{{\rm{e}}^4}}&0&{ - 4{{\rm{e}}^8}}&{{\rm{saddle{\ } point}}}
\end{array}$
