Answer
$$\left( { - 2,2} \right),\,\,\,\left( {0,0} \right),\,\,\,\left( {2,2} \right)$$
Work Step by Step
\[\begin{gathered}
f\left( {x,y} \right) = {x^4} + {y^4} - 16xy \hfill \\
{\text{Calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \hfill \\
{f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^4} + {y^4} - 16xy} \right] \hfill \\
{f_x}\left( {x,y} \right) = 4{x^3} - 16y \hfill \\
and \hfill \\
{f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^4} + {y^4} - 16xy} \right] \hfill \\
{f_y}\left( {x,y} \right) = 4{y^3} - 16x \hfill \\
\hfill \\
{\text{Set }}{f_x}\left( {x,y} \right) = 0{\text{ and }}{f_y}\left( {x,y} \right) = 0 \hfill \\
\left\{ {\begin{array}{*{20}{c}}
{4{x^3} - 16y = 0} \\
{4{y^3} - 16x = 0}
\end{array}} \right. \hfill \\
{\text{By symmetry }}x = y,{\text{ then}} \hfill \\
4{x^3} - 16x = 0 \hfill \\
4x\left( {{x^2} - 4} \right) = 0 \hfill \\
{x_1} = 0,\,\,\,{x_2} = 2,\,\,\,{x_3} = - 2 \hfill \\
{\text{The critical points occur at}} \hfill \\
\left( { - 2,2} \right),\,\,\,\left( {0,0} \right),\,\,\,\left( {2,2} \right) \hfill \\
\end{gathered} \]
