Answer
The critical points and their nature:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{}\\
{{\rm{Point}}}&{{\rm{Type}}}\\
{\left( {\frac{1}{{{\rm{e}}\sqrt 2 }},\frac{1}{{{\rm{e}}\sqrt 2 }}} \right)}&{{\rm{local{\ }minimum}}}\\
{\left( { - \frac{1}{{{\rm{e}}\sqrt 2 }}, - \frac{1}{{{\rm{e}}\sqrt 2 }}} \right)}&{{\rm{local{\ }maximum}}}\\
{\left( {\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right)}&{{\rm{saddle{\ }point}}}\\
{\left( { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)}&{{\rm{saddle{\ }point}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = \left( {x + y} \right)\ln \left( {{x^2} + {y^2}} \right)$.
The partial derivatives are
${f_x} = \ln \left( {{x^2} + {y^2}} \right) + \frac{{2x\left( {x + y} \right)}}{{{x^2} + {y^2}}}$, ${\ \ }$ ${f_y} = \ln \left( {{x^2} + {y^2}} \right) + \frac{{2y\left( {x + y} \right)}}{{{x^2} + {y^2}}}$
${f_{xx}} = \frac{{2x}}{{{x^2} + {y^2}}} + \frac{{\left( {{x^2} + {y^2}} \right)\left( {4x + 2y} \right) - 4{x^2}\left( {x + y} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
$ = \frac{{2{x^3} - 2{x^2}y + 6x{y^2} + 2{y^3}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
${f_{yy}} = \frac{{2y}}{{{x^2} + {y^2}}} + \frac{{\left( {{x^2} + {y^2}} \right)\left( {2x + 4y} \right) - 4{y^2}\left( {x + y} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
$ = \frac{{2{x^3} + 6{x^2}y - 2x{y^2} + 2{y^3}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
${f_{xy}} = \frac{{2y}}{{{x^2} + {y^2}}} + \frac{{2x\left( {{x^2} + {y^2}} \right) - 4xy\left( {x + y} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
$ = \frac{{2{{\left( {x - y} \right)}^2}\left( {x + y} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = \ln \left( {{x^2} + {y^2}} \right) + \frac{{2x\left( {x + y} \right)}}{{{x^2} + {y^2}}} = 0$
${f_y} = \ln \left( {{x^2} + {y^2}} \right) + \frac{{2y\left( {x + y} \right)}}{{{x^2} + {y^2}}} = 0$
From the first equation we obtain
(!) ${\ \ \ }$ $\ln \left( {{x^2} + {y^2}} \right) = - \frac{{2x\left( {x + y} \right)}}{{{x^2} + {y^2}}}$
Substituting it in the second equation gives
$ - \frac{{2x\left( {x + y} \right)}}{{{x^2} + {y^2}}} + \frac{{2y\left( {x + y} \right)}}{{{x^2} + {y^2}}} = 0$
$\frac{{ - 2{x^2} - 2xy + 2yx + 2{y^2}}}{{{x^2} + {y^2}}} = 0$
So, we solve the equation $ - 2{x^2} + 2{y^2} = 0$. The solution is $y = \pm x$.
1. Substituting $y=x$ in equation (1) gives
$\ln \left( {2{x^2}} \right) = - \frac{{4{x^2}}}{{2{x^2}}} = - 2$
${{\rm{e}}^{ - 2}} = 2{x^2}$, ${\ \ \ }$ $x = \pm \frac{1}{{{\rm{e}}\sqrt 2 }}$
The solutions are $\left( {\frac{1}{{{\rm{e}}\sqrt 2 }},\frac{1}{{{\rm{e}}\sqrt 2 }}} \right)$ and $\left( { - \frac{1}{{{\rm{e}}\sqrt 2 }}, - \frac{1}{{{\rm{e}}\sqrt 2 }}} \right)$.
2. Substituting $y=-x$ in equation (1) gives
$\ln \left( {2{x^2}} \right) = 0$
$2{x^2} = 1$, ${\ \ \ }$ $x = \pm \frac{1}{{\sqrt 2 }}$
The solutions are $\left( {\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right)$ and $\left( { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$.
So, the critical points are $\left( {\frac{1}{{{\rm{e}}\sqrt 2 }},\frac{1}{{{\rm{e}}\sqrt 2 }}} \right)$,$\left( { - \frac{1}{{{\rm{e}}\sqrt 2 }}, - \frac{1}{{{\rm{e}}\sqrt 2 }}} \right)$,$\left( {\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right)$ and $\left( { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$.
Recall from previous results:
${f_{xx}} = \frac{{2{x^3} - 2{x^2}y + 6x{y^2} + 2{y^3}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
${f_{yy}} = \frac{{2{x^3} + 6{x^2}y - 2x{y^2} + 2{y^3}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
${f_{xy}} = \frac{{2{{\left( {x - y} \right)}^2}\left( {x + y} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
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 it. The results are listed in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{}&{}&{}&{{\rm{Discriminant}}}&{}\\
{{\rm{Point}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\rm{Type}}}\\
{\left( {\frac{1}{{{\rm{e}}\sqrt 2 }},\frac{1}{{{\rm{e}}\sqrt 2 }}} \right)}&{2\sqrt 2 {\rm{e}}}&{2\sqrt 2 {\rm{e}}}&0&{8{{\rm{e}}^2}}&{{\rm{local{\ }minimum}}}\\
{\left( { - \frac{1}{{{\rm{e}}\sqrt 2 }}, - \frac{1}{{{\rm{e}}\sqrt 2 }}} \right)}&{ - 2\sqrt 2 {\rm{e}}}&{ - 2\sqrt 2 {\rm{e}}}&0&{8{{\rm{e}}^2}}&{{\rm{local{\ }maximum}}}\\
{\left( {\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right)}&{2\sqrt 2 }&{ - 2\sqrt 2 }&0&{ - 8}&{{\rm{saddle{\ }point}}}\\
{\left( { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)}&{ - 2\sqrt 2 }&{2\sqrt 2 }&0&{ - 8}&{{\rm{saddle{\ }point}}}
\end{array}$
