Answer
The critical points and their nature:
$\begin{array}{*{20}{c}}
{{\rm{Critical Point}}}&{{\rm{Type}}}\\
{\left( {0,0} \right)}&{{\rm{saddle{\ } point}}}\\
{\left( {\sqrt {\frac{1}{2}} ,\sqrt {\frac{1}{2}} } \right)}&{{\rm{local{\ } maximum}}}\\
{\left( {\sqrt {\frac{1}{2}} , - \sqrt {\frac{1}{2}} } \right)}&{{\rm{local{\ } minimum}}}\\
{\left( { - \sqrt {\frac{1}{2}} ,\sqrt {\frac{1}{2}} } \right)}&{{\rm{local{\ } minimum}}}\\
{\left( { - \sqrt {\frac{1}{2}} , - \sqrt {\frac{1}{2}} } \right)}&{{\rm{local{\ } maximum}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = xy{{\rm{e}}^{ - {x^2} - {y^2}}}$.
The partial derivatives are
${f_x} = y{{\rm{e}}^{ - {x^2} - {y^2}}} - 2{x^2}y{{\rm{e}}^{ - {x^2} - {y^2}}}$
$ = y\left( {1 - 2{x^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
${f_y} = x{{\rm{e}}^{ - {x^2} - {y^2}}} - 2x{y^2}{{\rm{e}}^{ - {x^2} - {y^2}}}$
$ = x\left( {1 - 2{y^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
${f_{xx}} = - 4xy{{\rm{e}}^{ - {x^2} - {y^2}}} - 2xy\left( {1 - 2{x^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
$ = \left( { - 4xy - 2xy\left( {1 - 2{x^2}} \right)} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
$ = \left( { - 6xy + 4{x^3}y} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
${f_{yy}} = - 4xy{{\rm{e}}^{ - {x^2} - {y^2}}} - 2xy\left( {1 - 2{y^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
$ = \left( { - 4xy - 2xy\left( {1 - 2{y^2}} \right)} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
$ = \left( { - 6xy + 4x{y^3}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
${f_{xy}} = \left( {1 - 2{x^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}} - 2{y^2}\left( {1 - 2{x^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
$ = \left( {1 - 2{x^2}} \right)\left( {1 - 2{y^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = y\left( {1 - 2{x^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}} = 0$
${f_y} = x\left( {1 - 2{y^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}} = 0$
Since ${{\rm{e}}^{ - {x^2} - {y^2}}} \ne 0$, the solutions of the first equation are $y=0$, $x = \pm \sqrt {\frac{1}{2}} $.
Substituting $y=0$ in the second equation gives $x=0$.
Substituting $x = \sqrt {\frac{1}{2}} $ in the second equation gives $y = \pm \sqrt {\frac{1}{2}} $.
Substituting $x = - \sqrt {\frac{1}{2}} $ in the second equation gives $y = \pm \sqrt {\frac{1}{2}} $.
So, the critical points are $\left( {0,0} \right)$, $\left( {\sqrt {\frac{1}{2}} ,\sqrt {\frac{1}{2}} } \right)$, $\left( {\sqrt {\frac{1}{2}} , - \sqrt {\frac{1}{2}} } \right)$, $\left( { - \sqrt {\frac{1}{2}} ,\sqrt {\frac{1}{2}} } \right)$, $\left( { - \sqrt {\frac{1}{2}} , - \sqrt {\frac{1}{2}} } \right)$.
Recall from previous results:
${f_{xx}} = \left( { - 6xy + 4{x^3}y} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
${f_{yy}} = \left( { - 6xy + 4x{y^3}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
${f_{xy}} = \left( {1 - 2{x^2}} \right)\left( {1 - 2{y^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
We evaluate ${f_{xx}}$, ${f_{yy}}$ and ${f_{xy}}$ at the corresponding critical points and use the Second Derivative Test to determine the nature of the critical points. The results are listed in the following table:
$\begin{array}{*{20}{c}}
{{\rm{CriticalPoint}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\rm{Type}}}\\
{\left( {0,0} \right)}&0&0&1&{ - 1}&{{\rm{saddle{\ } point}}}\\
{\left( {\sqrt {\frac{1}{2}} ,\sqrt {\frac{1}{2}} } \right)}&{ - \frac{2}{{\rm{e}}}}&{ - \frac{2}{{\rm{e}}}}&0&{\frac{4}{{{{\rm{e}}^2}}}}&{{\rm{local{\ } maximum}}}\\
{\left( {\sqrt {\frac{1}{2}} , - \sqrt {\frac{1}{2}} } \right)}&{\frac{2}{{\rm{e}}}}&{\frac{2}{{\rm{e}}}}&0&{\frac{4}{{{{\rm{e}}^2}}}}&{{\rm{local{\ } minimum}}}\\
{\left( { - \sqrt {\frac{1}{2}} ,\sqrt {\frac{1}{2}} } \right)}&{\frac{2}{{\rm{e}}}}&{\frac{2}{{\rm{e}}}}&0&{\frac{4}{{{{\rm{e}}^2}}}}&{{\rm{local{\ } minimum}}}\\
{\left( { - \sqrt {\frac{1}{2}} , - \sqrt {\frac{1}{2}} } \right)}&{ - \frac{2}{{\rm{e}}}}&{ - \frac{2}{{\rm{e}}}}&0&{\frac{4}{{{{\rm{e}}^2}}}}&{{\rm{local{\ } maximum}}}
\end{array}$
