Answer
$\begin{array}{*{20}{c}}
{{\bf{Critical}}}&{}\\
{{\bf{Point}}}&{{\bf{Type}}}\\
{\left( {0,0} \right)}&{{\rm{saddle{\ }point}}}\\
{\left( {\frac{6}{{{{108}^{2/3}}}},\frac{1}{{{{108}^{1/3}}}}} \right) \simeq \left( {0.26,0.21} \right)}&{{\rm{local{\ }minimum}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = {x^3} + 2{y^3} - xy$.
The partial derivatives are
${f_x} = 3{x^2} - y$, ${\ \ \ }$ ${f_y} = 6{y^2} - x$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 3{x^2} - y = 0$
$y = 3{x^2}$
${f_y} = 6{y^2} - x = 0$
$x = 6{y^2}$
Substituting $x = 6{y^2}$ in $y = 3{x^2}$ gives
$y = 3{\left( {6{y^2}} \right)^2}$
$108{y^4} - y = 0$
$y\left( {108{y^3} - 1} \right) = 0$
$y = 0$, ${\ \ \ \ }$ $y = \frac{1}{{{{108}^{1/3}}}}$
Using $x = 6{y^2}$, we obtain the critical points: $\left( {0,0} \right)$ and $\left( {\frac{6}{{{{108}^{2/3}}}},\frac{1}{{{{108}^{1/3}}}}} \right) \simeq \left( {0.26,0.21} \right)$.
The Second Derivative Test:
We evaluate the second partial derivatives:
${f_{xx}} = 6x$, ${\ \ \ }$ ${f_{yy}} = 12y$, ${\ \ \ }$ ${f_{xy}} = - 1$
Using the Second Derivative Test, we determine the nature of the critical points and list the results in the following table:
$\begin{array}{*{20}{c}}
{{\bf{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\bf{Discriminant}}}&{}\\
{{\bf{Point}}}&{6x}&{12y}&{ - 1}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\bf{Type}}}\\
{\left( {0,0} \right)}&0&0&{ - 1}&{ - 1}&{{\rm{saddle{\ }point}}}\\
{\left( {\frac{6}{{{{108}^{2/3}}}},\frac{1}{{{{108}^{1/3}}}}} \right)}&{\frac{{36}}{{{{108}^{2/3}}}}}&{\frac{{12}}{{{{108}^{1/3}}}}}&{ - 1}&3&{{\rm{local{\ }minimum}}}
\end{array}$
