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