Answer
The critical points and their nature:
$\begin{array}{*{20}{c}}
{{\bf{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\bf{Discriminant}}}&{}\\
{{\bf{Point}}}&{ - 18x}&{ - 4x}&{ - 4y}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\bf{Type}}}\\
{\left( {0,\sqrt 2 } \right)}&0&0&{ - 4\sqrt 2 }&{ - 32}&{{\rm{saddle{\ }point}}}\\
{\left( {0, - \sqrt 2 } \right)}&0&0&{4\sqrt 2 }&{ - 32}&{{\rm{saddle{\ }point}}}\\
{\left( {\frac{2}{3},0} \right)}&{ - 12}&{ - \frac{8}{3}}&0&{32}&{{\rm{local{\ }maximum}}}\\
{\left( { - \frac{2}{3},0} \right)}&{12}&{\frac{8}{3}}&0&{32}&{{\rm{local{\ }minimum}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = 4x - 3{x^3} - 2x{y^2}$.
The partial derivatives are
${f_x} = 4 - 9{x^2} - 2{y^2}$, ${\ \ \ }$ ${f_y} = - 4xy$
${f_{xx}} = - 18x$, ${\ \ }$ ${f_{yy}} = - 4x$, ${\ \ }$ ${f_{xy}} = - 4y$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 4 - 9{x^2} - 2{y^2} = 0$, ${\ \ }$ ${f_y} = - 4xy = 0$
From the second equation we obtain the solutions $x=0$ or $y=0$.
For $x=0$, we obtain from the first equation $y = \pm \sqrt 2 $.
For $y=0$, we obtain from the first equation $x = \pm \frac{2}{3}$.
So, the critical points are $\left( {0,\sqrt 2 } \right)$, $\left( {0, - \sqrt 2 } \right)$, $\left( {\frac{2}{3},0} \right)$ and $\left( { - \frac{2}{3},0} \right)$.
Next, we use the Second Derivative Test to determine the nature of the critical points. The results are listed in the following table:
$\begin{array}{*{20}{c}}
{{\bf{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\bf{Discriminant}}}&{}\\
{{\bf{Point}}}&{ - 18x}&{ - 4x}&{ - 4y}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\bf{Type}}}\\
{\left( {0,\sqrt 2 } \right)}&0&0&{ - 4\sqrt 2 }&{ - 32}&{{\rm{saddle{\ }point}}}\\
{\left( {0, - \sqrt 2 } \right)}&0&0&{4\sqrt 2 }&{ - 32}&{{\rm{saddle{\ }point}}}\\
{\left( {\frac{2}{3},0} \right)}&{ - 12}&{ - \frac{8}{3}}&0&{32}&{{\rm{local{\ }maximum}}}\\
{\left( { - \frac{2}{3},0} \right)}&{12}&{\frac{8}{3}}&0&{32}&{{\rm{local{\ }minimum}}}
\end{array}$
