Answer
Local minimum of $f(-3,3)=-5$
Work Step by Step
Since, we have $f_x(x,y)=2x+y+3=0, f_y(x,y)=x+2y-3=0$
After solving the above two equations, we get
$x=-3,y=3$
Thus the critical point is: $(-3,3)$
As per second derivative test, we have
$D=f_{xx}f_{yy}-f^2_{xy}=3$ and $D=3 \gt 0$ and $f_{xx}(-3,3)=2 \gt 0$
Thus, local minimum of $f(-3,3)=-5$
