= Solution
The <determinant> of the <Hessian matrix> is $4\cdot2-3^2=-1<0$. Hence it has one positive and one negative <eigenvalue> (explicitly $3\pm\sqrt{10}$), so its quadratic form is indefinite. The cubic terms are of higher order and cannot remove these opposite local directions of change. Therefore
$$
\boxed{(0,0)\text{ is a saddle point}.}
$$
For example the quadratic term is positive along $y=0$ but negative along $y=-3x/2$, confirming the <saddle point of a scalar function> classification.
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