Answer
$\begin{array}{*{20}{c}}
{{\bf{Critical}}}&{}\\
{{\bf{Point}}}&{{\bf{Type}}}\\
{\left( {0,0} \right)}&{{\rm{saddle{\ }point}}}\\
{\left( {1,1} \right)}&{{\rm{local{\ }minimum}}}\\
{\left( { - 1, - 1} \right)}&{{\rm{local{\ }minimum}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = {x^4} - 4xy + 2{y^2}$.
The partial derivatives are
${f_x} = 4{x^3} - 4y$, ${\ \ \ }$ ${f_y} = 4y - 4x$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 4{x^3} - 4y = 0$
$y = {x^3}$
${f_y} = 4y - 4x = 0$
$y = x$
From these equations we obtain $x = {x^3}$.
${x^3} - x = 0$
$x\left( {{x^2} - 1} \right) = 0$
$x = 0$, ${\ \ \ \ }$ $x = \pm 1$
Using $y = x$, the critical points are $\left( {0,0} \right)$, $\left( {1,1} \right)$ and $\left( { - 1, - 1} \right)$.
The Second Derivative Test:
We evaluate the second partial derivatives:
${f_{xx}} = 12{x^2}$, ${\ \ \ }$ ${f_{yy}} = 4$, ${\ \ \ }$ ${f_{xy}} = - 4$
Using the Second Derivative Test, we determine the nature of the critical points and list the results in the following table:
$\begin{array}{*{20}{c}}
{{\bf{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\bf{Discriminant}}}&{}\\
{{\bf{Point}}}&{12{x^2}}&4&{ - 4}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\bf{Type}}}\\
{\left( {0,0} \right)}&0&4&{ - 4}&{ - 16}&{{\rm{saddle{\ }point}}}\\
{\left( {1,1} \right)}&{12}&4&{ - 4}&{32}&{{\rm{local{\ }minimum}}}\\
{\left( { - 1, - 1} \right)}&{12}&4&{ - 4}&{32}&{{\rm{local{\ }minimum}}}
\end{array}$
