Solution
= Solution
The <gradient> is $(2x+2,-2y-4)$, so the only <stationary point> is $(x,y)=(-1,-2)$. Its <Hessian matrix> is $\operatorname{diag}(2,-2)$ and its determinant is $-4<0$. Thus
$$
\boxed{(x,y)=(-1,-2),\qquad f=13,\qquad\text{saddle point}.}
$$
Completing squares confirms the classification directly: $f=(x+1)^2-(y+2)^2+13$. Arbitrarily close to the point, moving in the $x$ direction increases the value and moving in the $y$ direction decreases it.