Answer
The critical point and its nature:
$\begin{array}{*{20}{c}}
{{\bf{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\bf{Discriminant}}}&{}\\
{{\bf{Point}}}&{6xy + 24}&0&{3{x^2}}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\bf{Type}}}\\
{\left( {2, - 4} \right)}&{ - 24}&0&{12}&{ - 144}&{{\rm{saddle{\ }point}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = {x^3}y + 12{x^2} - 8y$.
The partial derivatives are
${f_x} = 3{x^2}y + 24x$, ${\ \ \ }$ ${f_y} = {x^3} - 8$
${f_{xx}} = 6xy + 24$, ${\ \ }$ ${f_{yy}} = 0$, ${\ \ }$ ${f_{xy}} = 3{x^2}$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 3{x^2}y + 24x = 0$, ${\ \ }$ ${f_y} = {x^3} - 8 = 0$
From the second equation we get $x=2$. Substituting it in the first equation gives
$12y + 48 = 0$, ${\ \ \ }$ $y = - 4$
So, there is only one critical point: $\left( {2, - 4} \right)$.
Next, we use the Second Derivative Test to determine the nature of the critical point. The results are listed in the following table:
$\begin{array}{*{20}{c}}
{{\bf{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\bf{Discriminant}}}&{}\\
{{\bf{Point}}}&{6xy + 24}&0&{3{x^2}}&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\bf{Type}}}\\
{\left( {2, - 4} \right)}&{ - 24}&0&{12}&{ - 144}&{{\rm{saddle{\ }point}}}
\end{array}$
