Answer
The critical points and their nature:
$\begin{array}{*{20}{c}}
{{\bf{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\bf{Discriminant}}}&{}\\
{{\bf{Point}}}&{6x}&{6y}&{ - 1}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\bf{Type}}}\\
{\left( {0,0} \right)}&0&0&{ - 1}&{ - 1}&{{\rm{saddle{\ }point}}}\\
{\left( {\frac{1}{3},\frac{1}{3}} \right)}&2&2&{ - 1}&3&{{\rm{local{\ }minimum}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = {x^3} - xy + {y^3}$.
The partial derivatives are
${f_x} = 3{x^2} - y$, ${\ \ \ }$ ${f_y} = - x + 3{y^2}$
${f_{xx}} = 6x$, ${\ \ }$ ${f_{yy}} = 6y$, ${\ \ }$ ${f_{xy}} = - 1$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 3{x^2} - y = 0$, ${\ \ \ }$ ${f_y} = - x + 3{y^2} = 0$
From the first equation we get $y = 3{x^2}$. Substituting it in the second equation gives
$ - x + 27{x^4} = 0$
$x\left( {27{x^3} - 1} \right) = 0$
The solutions are $x = 0$, $x = \frac{1}{3}$.
So, there are two critical points: $\left( {0,0} \right)$ and $\left( {\frac{1}{3},\frac{1}{3}} \right)$.
Next, we use the Second Derivative Test to determine the nature of the critical points. The results are listed in the following table:
$\begin{array}{*{20}{c}}
{{\bf{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\bf{Discriminant}}}&{}\\
{{\bf{Point}}}&{6x}&{6y}&{ - 1}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\bf{Type}}}\\
{\left( {0,0} \right)}&0&0&{ - 1}&{ - 1}&{{\rm{saddle{\ }point}}}\\
{\left( {\frac{1}{3},\frac{1}{3}} \right)}&2&2&{ - 1}&3&{{\rm{local{\ }minimum}}}
\end{array}$
