Thomas' Calculus 13th Edition

Saddle Point of $f(0,-1)=2$
$$f_x(x,y)=10=4y+4=0$$ and $$f_y(x,y)=4x-4y-4=0$$ Simplify the above two equations to find the critical point at $(0,-1)$. We need to apply the second derivative test for $(0,-1)$ $D=f_{xx}f_{yy}-f^2_{xy}=(10)(-4)-(4)^2=-56 \lt 0$ This implies that we have a Saddle Point of $f(0,-1)=2$