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