Answer
There is a relative minimum at $(4,-2)$
Work Step by Step
$f_{x}(x,y)=2x + y - 6 = 0$
$f_{y}(x,y)=x + 2y=0 \rightarrow x=-2y$
Use the substitution method to solve the system of equations
$2(-2y)+y-6=0 \rightarrow y=-2$
$x=4$
The critical points is $(4,-2)$
$f_{xx}(x,y)=2$
$f_{yy}(x,y)=2$
$f_{xy}(x,y)=1$
$f_{xx}(4,-2)=2$
$f_{yy}(4,-2)=2$
$f_{xy}(4,-2)=1$
$D=2*2-(1)^{2}=3$
Since there is D>0, and $f_{xx}(x,y)=2 \gt 0$, there is a relative minimum at $(4,-2)$
