Answer
The critical point and its nature:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{}\\
{{\rm{Point}}}&{{\rm{Type}}}\\
{\left( { - \frac{1}{6}, - \frac{{17}}{{18}}} \right)}&{{\rm{local{\ }minimum}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = \left( {x + 3y} \right){{\rm{e}}^{y - {x^2}}}$.
The partial derivatives are
${f_x} = {{\rm{e}}^{y - {x^2}}} - 2x\left( {x + 3y} \right){{\rm{e}}^{y - {x^2}}}$
$ = \left( { - 2{x^2} - 6xy + 1} \right){{\rm{e}}^{y - {x^2}}}$
${f_y} = 3{{\rm{e}}^{y - {x^2}}} + \left( {x + 3y} \right){{\rm{e}}^{y - {x^2}}}$
$ = \left( {x + 3y + 3} \right){{\rm{e}}^{y - {x^2}}}$
${f_{xx}} = \left( { - 4x - 6y} \right){{\rm{e}}^{y - {x^2}}} - 2x\left( { - 2{x^2} - 6xy + 1} \right){{\rm{e}}^{y - {x^2}}}$
$ = 2{{\rm{e}}^{ - {x^2} + y}}\left( {2{x^3} + 6{x^2}y - 3x - 3y} \right)$
${f_{yy}} = 3{{\rm{e}}^{y - {x^2}}} + \left( {x + 3y + 3} \right){{\rm{e}}^{y - {x^2}}}$
$ = \left( {x + 3y + 6} \right){{\rm{e}}^{y - {x^2}}}$
${f_{xy}} = - 6x{{\rm{e}}^{y - {x^2}}} + \left( { - 2{x^2} - 6xy + 1} \right){{\rm{e}}^{y - {x^2}}}$
$ = \left( { - 2{x^2} - 6xy - 6x + 1} \right){{\rm{e}}^{y - {x^2}}}$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = \left( { - 2{x^2} - 6xy + 1} \right){{\rm{e}}^{y - {x^2}}} = 0$
${f_y} = \left( {x + 3y + 3} \right){{\rm{e}}^{y - {x^2}}} = 0$
Since ${{\rm{e}}^{y - {x^2}}} \ne 0$, we obtain two simultaneous equations:
(1) ${\ \ \ }$ $ - 2{x^2} - 6xy + 1 = 0$ ${\ }$ and ${\ }$ $x + 3y + 3 = 0$
The second equation gives $y = - \frac{{x + 3}}{3}$.
Substituting it in the first equation gives
$ - 2{x^2} - 6x\left( { - \frac{{x + 3}}{3}} \right) + 1 = 0$
$ - 2{x^2} + 2x\left( {x + 3} \right) + 1 = 0$
$6x + 1 = 0$, ${\ \ \ }$ $x = - \frac{1}{6}$
Substituting $x = - \frac{1}{6}$ in $y = - \frac{{x + 3}}{3}$ gives $y = - \frac{{17}}{{18}}$.
So, there is only one critical point: $\left( { - \frac{1}{6}, - \frac{{17}}{{18}}} \right)$.
We evaluate ${f_{xx}}$, ${f_{yy}}$ and ${f_{xy}}$ at the corresponding critical point and use the Second Derivative Test to determine the nature of the critical point. The results are listed in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{}&{}&{}&{{\rm{Discriminant}}}&{}\\
{{\rm{Point}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\rm{Type}}}\\
{\left( { - \frac{1}{6}, - \frac{{17}}{{18}}} \right)}&{2.396}&{1.135}&{0.378}&{2.575}&{{\rm{local{\ }minimum}}}
\end{array}$
