= Solution
The <gradient> equations are
$$
f_x=3x^2+2xy-10x-9=0,\qquad f_y=x^2-9=0.
$$
Thus $x=3$ gives $y=2$, while $x=-3$ gives $y=8$. The <Hessian matrix> is
$$
H=\begin{pmatrix}6x+2y-10&2x\\2x&0\end{pmatrix},\qquad\det H=-4x^2.
$$
At both critical points the determinant is $-36<0$, so the Hessian is indefinite and the <second-derivative test> classifies both as <saddle points>. Direct substitution gives
$$
\boxed{(3,2),\ (-3,8)\text{ are saddle points, and }f(3,2)=f(-3,8)=-30.}
$$
The factorization $f+30=(x^2-9)(x+y-5)$ also shows the crossing level-set branches through each saddle.
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