How do you find the critical point and determine whether

${\displaystyle f(x,y)=x^2+4x+y^2}$
${\displaystyle f_x(x,y)=2x+4}$
${\displaystyle f_y(x,y)=2y}$
The critical point makes both partial derivatives 0 (simultaneously).
For this function there is one critical point: ${\displaystyle (-2,0)}$
To determine whether f has a local minimum, maximum or neither at this point we apply the second derivative test for functions of two variables. (Well, we try to apply it. It does not always give an answer.)
${\displaystyle f_{xx}(x,y)=2}$
${\displaystyle f_{xy}(x,y)=0}$ (As usual, this is also ${\displaystyle f_{yx}(x,y)}$)
${\displaystyle f_{yy}(x,y)=2}$
Evaluate the second partials at the critical point (In this case they are all constant, but in general we cannot skip this step.)
At the critical point ${\displaystyle (-2,0)}$, we get
${\displaystyle A=f_{xx}(-2,0)=2}$
${\displaystyle B=f_{xy}(-2,0)=0}$
${\displaystyle C=f_{yy}(-2,0)=2}$
Calculate ${\displaystyle D=AC-B^2}$
${\displaystyle D=\left(2\right)\left(2\right)-{\left(0\right)}^2=4}$
Apply the second derivative test:
Since D is positive, we look at A and with ${\displaystyle D>0}$ and ${\displaystyle A>0}$, we have a local minimum at the critical point.
f ${\displaystyle (-2,0)=4-8=-4}$
f has a local minimum of−4 at ${\displaystyle (-2,0)}$.