Classification of critical points with the Hessian matrix

The gradient vanishes only at $(2,-1)$, hence by Fermat’s theorem any unconstrained extremum is at that point. The second derivatives are constant: $f_{xx}=2$, $f_{yy}=2$, $f_{xy}=0$. Since $\det H=4>0$ and $f_{xx}=2>0$, the Hessian matrix is positive definite, therefore $(2,-1)$ is a local minimum point with value $f(2,-1)=-5$. The minimum is also absolute, since completing the squares gives $f(x,y)=(x-2{)}^2+(y+1{)}^2-5\ge -5$ for every $(x,y)$.

The first equation of the system gives $x^2=1$, that is $x=\pm 1$, and the second gives $y=0$, hence there are two critical points. The Hessian matrix depends only on $x$ and must be evaluated at each point. At $(1,0)$ the determinant is positive and $f_{xx}=6>0$, therefore the point is a local minimum point with $f(1,0)=1-3=-2$. At $(-1,0)$ the determinant is negative, therefore the point is a saddle point. The minimum is not absolute, since along the $x$-axis we have $f(x,0)=x^3-3x\to -\infty$ as $x\to -\infty$.

The system is not decoupled, hence we proceed by substitution: from the first equation we obtain $y=\frac{x^2}{2}$ and substitute it into the second. The resulting equation factors as $\frac{3x}{4}(x^3-8)=0$ and has only the real solutions $x=0$ and $x=2$; the corresponding values $y=\frac{x^2}{2}$ give the critical points $(0,0)$ and $(2,2)$. At the origin the determinant of the Hessian is negative, therefore the point is a saddle point. At $(2,2)$ the determinant is positive and $f_{xx}=12>0$, therefore the point is a local minimum point with $f(2,2)=-8$. Here too the minimum is not absolute: along the $x$-axis we have $f(x,0)=x^3\to -\infty$ as $x\to -\infty$.