 Solution

The determinant of the Hessian matrix is $4\cdot 2-3^2=-1<0$. 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. ThereforeFor example the quadratic term is positive along $y=0$ but negative along $y=-3x/2$, confirming the saddle point of a scalar function classification.