Answer
Using a computer algebra system we obtain the critical point at $\left( {x,y} \right) = \left( {0.278,0.278} \right)$.
Applying the Second Derivative Test, it is confirmed that the critical point $\left( {0.278,0.278} \right)$ corresponds to a local minimum as in Figure 22.
Work Step by Step
We have $f\left( {x,y} \right) = \left( {1 - x + {x^2}} \right){{\rm{e}}^{{y^2}}} + \left( {1 - y + {y^2}} \right){{\rm{e}}^{{x^2}}}$.
The partial derivatives are
${f_x} = \left( { - 1 + 2x} \right){{\rm{e}}^{{y^2}}} + 2x\left( {1 - y + {y^2}} \right){{\rm{e}}^{{x^2}}}$
${f_y} = 2y\left( {1 - x + {x^2}} \right){{\rm{e}}^{{y^2}}} + \left( { - 1 + 2y} \right){{\rm{e}}^{{x^2}}}$
Using a computer algebra system we solve the simultaneous equations above and obtain the critical point at $\left( {x,y} \right) = \left( {0.278,0.278} \right)$.
The second partial derivatives are
${f_{xx}} = 2{{\rm{e}}^{{y^2}}} + 2{{\rm{e}}^{{x^2}}}\left( {1 + 2{x^2}} \right)\left( {1 - y + {y^2}} \right)$
${f_{yy}} = 2{{\rm{e}}^{{x^2}}} + 2{{\rm{e}}^{{y^2}}}\left( {1 - x + {x^2}} \right)\left( {1 + 2{y^2}} \right)$
${f_{xy}} = 2y\left( { - 1 + 2x} \right){{\rm{e}}^{{y^2}}} + 2x\left( { - 1 + 2y} \right){{\rm{e}}^{{x^2}}}$
We evaluate ${f_{xx}}$, ${f_{yy}}$ and ${f_{xy}}$ at the critical point $\left( {x,y} \right) = \left( {0.278,0.278} \right)$ and apply the Second Derivative Test to determine the nature of the critical point. Since $D>0$ and ${f_{xx}} > 0$, it is confirmed that the critical point $\left( {0.278,0.278} \right)$ corresponds to a local minimum as in Figure 22. The results are listed in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{}&{}&{}&{{\rm{Discriminant}}}&{}\\
{{\rm{Point}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\rm{Type}}}\\
{\left( {0.278,0.278} \right)}&{4.155}&{4.155}&{ - 0.533}&{16.977}&{{\rm{local{\ }minimum}}}
\end{array}$
