Answer
The critical points and their nature:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}\\
{{\rm{Point}}}&{6x}&{12{y^2} - 4}&0\\
{\left( {\sqrt 2 ,0} \right)}&{6\sqrt 2 }&{ - 4}&0\\
{\left( { - \sqrt 2 ,0} \right)}&{ - 6\sqrt 2 }&{ - 4}&0\\
{\left( {\sqrt 2 ,1} \right)}&{6\sqrt 2 }&8&0\\
{\left( {\sqrt 2 , - 1} \right)}&{6\sqrt 2 }&8&0\\
{\left( { - \sqrt 2 ,1} \right)}&{ - 6\sqrt 2 }&8&0\\
{\left( { - \sqrt 2 , - 1} \right)}&{ - 6\sqrt 2 }&8&0
\end{array}\begin{array}{*{20}{c}}
{{\rm{Discriminant}}}&{}\\
{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\rm{Type}}}\\
{ - 24\sqrt 2 }&{{\rm{saddle{\ }point}}}\\
{24\sqrt 2 }&{{\rm{local{\ }maximum}}}\\
{48\sqrt 2 }&{{\rm{local{\ }minimum}}}\\
{48\sqrt 2 }&{{\rm{local{\ }minimum}}}\\
{ - 48\sqrt 2 }&{{\rm{saddle{\ }point}}}\\
{ - 48\sqrt 2 }&{{\rm{saddle{\ }point}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = {x^3} + {y^4} - 6x - 2{y^2}$.
The partial derivatives are
${f_x} = 3{x^2} - 6$, ${\ \ \ }$ ${f_y} = 4{y^3} - 4y$
${f_{xx}} = 6x$, ${\ \ }$ ${f_{yy}} = 12{y^2} - 4$, ${\ \ }$ ${f_{xy}} = 0$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 3{x^2} - 6 = 0$, ${\ \ \ }$ ${f_y} = 4{y^3} - 4y = 0$
From the first equation we obtain the solutions $x = \pm \sqrt 2 $.
From the second equation we have
$4y\left( {{y^2} - 1} \right) = 0$
The solutions are $y=0$, $y = \pm 1$.
So, the critical points are $\left( {\sqrt 2 ,0} \right)$, $\left( { - \sqrt 2 ,0} \right)$, $\left( {\sqrt 2 ,1} \right)$, $\left( {\sqrt 2 , - 1} \right)$, $\left( { - \sqrt 2 ,1} \right)$, $\left( { - \sqrt 2 , - 1} \right)$.
Next, we use the Second Derivative Test to determine the nature of the critical points. The results are listed in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}\\
{{\rm{Point}}}&{6x}&{12{y^2} - 4}&0\\
{\left( {\sqrt 2 ,0} \right)}&{6\sqrt 2 }&{ - 4}&0\\
{\left( { - \sqrt 2 ,0} \right)}&{ - 6\sqrt 2 }&{ - 4}&0\\
{\left( {\sqrt 2 ,1} \right)}&{6\sqrt 2 }&8&0\\
{\left( {\sqrt 2 , - 1} \right)}&{6\sqrt 2 }&8&0\\
{\left( { - \sqrt 2 ,1} \right)}&{ - 6\sqrt 2 }&8&0\\
{\left( { - \sqrt 2 , - 1} \right)}&{ - 6\sqrt 2 }&8&0
\end{array}\begin{array}{*{20}{c}}
{{\rm{Discriminant}}}&{}\\
{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\rm{Type}}}\\
{ - 24\sqrt 2 }&{{\rm{saddle{\ }point}}}\\
{24\sqrt 2 }&{{\rm{local{\ }maximum}}}\\
{48\sqrt 2 }&{{\rm{local{\ }minimum}}}\\
{48\sqrt 2 }&{{\rm{local{\ }minimum}}}\\
{ - 48\sqrt 2 }&{{\rm{saddle{\ }point}}}\\
{ - 48\sqrt 2 }&{{\rm{saddle{\ }point}}}
\end{array}$
