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