Answer
The critical points and their nature:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{}\\
{{\rm{Point}}}&{{\rm{Type}}}\\
{\left( {0,0} \right)}&{{\rm{saddle{\ } point}}}\\
{\left( {1,1} \right)}&{{\rm{local{\ } minimum}}}\\
{\left( { - 1, - 1} \right)}&{{\rm{local{\ } minimum}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = {x^4} + {y^4} - 4xy$.
The partial derivatives are
${f_x} = 4{x^3} - 4y$, ${\ \ \ }$ ${f_y} = 4{y^3} - 4x$
${f_{xx}} = 12{x^2}$, ${\ \ }$ ${f_{yy}} = 12{y^2}$, ${\ \ }$ ${f_{xy}} = - 4$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 4{x^3} - 4y = 0$, ${\ \ \ }$ ${f_y} = 4{y^3} - 4x = 0$
From the first equation we obtain $y = {x^3}$. Substituting it in the second equation gives
$4{x^9} - 4x = 0$
$4x\left( {{x^8} - 1} \right) = 0$
The solutions are $x=0$, $x = \pm 1$.
So, the critical points are $\left( {0,0} \right)$, $\left( {1,1} \right)$, $\left( { - 1, - 1} \right)$.
Next, we use the Second Derivative Test to determine the nature of the critical points. The results are listed in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\rm{Discriminant}}}&{}\\
{{\rm{Point}}}&{12{x^2}}&{12{y^2}}&{ - 4}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\rm{Type}}}\\
{\left( {0,0} \right)}&0&0&{ - 4}&{ - 16}&{{\rm{saddle{\ } point}}}\\
{\left( {1,1} \right)}&{12}&{12}&{ - 4}&{128}&{{\rm{local{\ } minimum}}}\\
{\left( { - 1, - 1} \right)}&{12}&{12}&{ - 4}&{128}&{{\rm{local{\ } minimum}}}
\end{array}$
