For $q_\pm=(\pm 1,0)$, ${\rm det\ Hess}\ f>0$ : Hence it is local minimum or local maximum : $f(1,0)=1$ and $f(1+\epsilon,0)=1 \- \frac{3}{2}\epsilon^2 - \frac{1}{2}\epsilon^3$ so that $q_+$ is local maximum
Since $f(-1,0)=-1$ and $f(-1-\epsilon,0)=-f(1+\epsilon,0)$ so it is local minimum
For $p_\pm = (0,\pm\sqrt{\frac{3}{2}})$, $ {\rm det\ Hess}\ f <0$ Hence at $p_+$, there are two curves $\alpha_i$ s.t. $f\circ \alpha_1$ has local minimum and $f\circ \alpha_2$ has local maximum