Answer
Saddle point at $f(-\dfrac{1}{2},\dfrac{3}{2})$
Work Step by Step
Given: $f_x(x,y)=2x+\dfrac{1}{x+y}=0, f_y(x,y)=-1+\dfrac{1}{x+y}=0$
Simplify the given two equations.
Critical point: $(-\dfrac{1}{2},\dfrac{3}{2})$
In order to solve this problem we will have to apply Second derivative test that suggests the following conditions to calculate the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$.
1. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\gt 0$ , then $f(a,b)$ is a local minimum.
2. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\lt 0$ , then $f(a,b)$ is a local maximum.
3. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \lt 0$ , then $f(a,b)$ is a not a local minimum or local maximum but a saddle point.
$D(\dfrac{1}{2},1)=f_{xx}f_{yy}-f^2_{xy}=-2 \implies D=-2 \lt 0$
Hence, Saddle point at $f(-\dfrac{1}{2},\dfrac{3}{2})$
