We determine the extrema of a function $z(x,y)=3x^2+2xy+\frac{1}{3}y^3$. We first have to find the stationary points. Hence, we have to find all the points $(x,y)$ such that the first-order partial derivatives $z'_x(x,y)$ and $z'_y(x,y)$ are both equal to zero. This gives the system
$$\left\{ \begin{array}{lr} z'_x(x,y)=6x+2y=0 & (1)\\z'_y(x,y)=2x+y^2=0 & (2)\end{array} \right.$$
To simplify this system we subtract equation 1 three times from equation 2. This gives the system
$$\left\{ \begin{array}{lr} 2y-3y^2=0& (1)\\2x+y^2=0& (2) \end{array} \right.$$
We can simplify this system by rewriting the equations to
$$\left\{ \begin{array}{lr} 2y(1-\frac{3}{2}y)=0& (1)\\x=-\frac{1}{2}y^2& (2) \end{array} \right.$$
From equation 1 it follows that $y=0$ or $y=\frac{2}{3}$. When we plug these values into equation 2, then we find that $x=0$ or $x=-\frac{2}{9}$. The stationary points of $z(x,y)$ are therefore $(x,y)=(0,0)$ and $(x,y)=(-\frac{2}{9},\frac{2}{3})$.
To determine whether these stationary points are extrema we have to plug these points into the criterion function. From $z''_{xx}(x,y)=6$, $z''_{yy}(x,y)=2y$ and $z''_{xy}(x,y)=z''_{yx}(x,y)=2$ it follows that this function is given by
$$C(x,y)=12y-4.$$
Plugging $(0,0)$ into this function gives $C(0,0)=-4<0$, which means that according to the second-order condition the point $(0,0)$ is a saddle point.
Plugging $(-\frac{2}{9},\frac{2}{3})$ into the criterion function gives $C(-\frac{2}{9},\frac{2}{3})=4>0$. Moreover, $z''_{xx}(-\frac{2}{9},\frac{2}{3})=6>0$, which means by the second-order condition that $(-\frac{2}{9},\frac{2}{3})$ is a minimum location.
