University Calculus: Early Transcendentals (3rd Edition)

Local minimum of $f(-3,3)=-5$
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 $f(-3,3)=-5$