Answer
Saddle point at $f(-\dfrac{1}{2},\dfrac{3}{2})$
Work Step by Step
Since, we have $f_x(x,y)=2x+\dfrac{1}{x+y}=0, f_y(x,y)=-1+\dfrac{1}{x+y}=0$
After solving the above two equations, we get
The critical point is: $(-\dfrac{1}{2},\dfrac{3}{2})$
As per the second derivative test, for $(\dfrac{1}{2},1)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=-2$ and $D=-2 \lt 0$
Thus, we have: Saddle point at $f(-\dfrac{1}{2},\dfrac{3}{2})$
