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