 Solution

Solving $f_y=0$ gives $y=-1-\alpha x$. Substitution into $f_x=0$ gives $x\{3x+2(1-{\alpha }^2)\}=0$. Therefore $(0,-1)$ is always a critical point. When $\alpha \ne -1$ as well as the printed restriction $\alpha \ne 1$, there is a second critical pointAt $(0,-1)$, the Hessian matrix is $\left(\begin{array}{cc}2& 2\alpha \\ 2\alpha & 2\end{array}\right)$ with determinant $4(1-{\alpha }^2)$. The second-derivative test gives a strict local minimum when $|\alpha |<1$ and a saddle point of a scalar function when $|\alpha |>1$.

The allowed value $\alpha =-1$ requires higher-order terms. With $v=y+1+\alpha x$, completing the square gives the exact identityThe coordinate change is invertible, so the Hessian at a critical point under a change of coordinates preserves the classification. At $\alpha =-1$, this is a cubic degenerate saddle: along $v=0$, the difference from the critical value is $x^3$ and takes both signs. Thus, throughout the printed parameter range,There is no local maximum. When the second point is distinct, it is a local minimum for $|\alpha |>1$ and a saddle point of a scalar function for $|\alpha |<1$: in the $(x,v)$ coordinates its Hessian matrix is $\mathrm{diag}(-2(1-{\alpha }^2),2)$. The printed instruction to split only at $\alpha =1$ misses the negative-parameter cases.