Example 1 (to $1^{\circ}$). Investigate the function $z=$ $=x^{3}+y^{3}-3 x y$ for extremum.
Solution. The function is defined and differentiable for all $(x, y)$.

1) Find the stationary points:

$$
\left\{\begin{array} { l } 
{ z _ { x } ^ { \prime } = 3 x ^ { 2 } - 3 y = 0 , } \\
{ z _ { y } ^ { \prime } = 3 y ^ { 2 } - 3 x = 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x ^ { 2 } - y = 0 , } \\
{ y ^ { 2 } - x = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y=x^{2} \\
x^{4}-x=0
\end{array}\right.\right.\right.
$$

We have two stationary points $M_{1}(1,1)$ and $M_{2}(0,0)$.

2) Check the sufficient conditions:

$$
z_{x^{2}}^{\prime \prime}=6 x, \quad z_{x y}^{\prime \prime}=-3, \quad z_{y^{2}}^{\prime \prime}=6 y
$$

For $M_{1}(1,1)$ we have:

$$
\begin{gathered}
A_{1}=z_{x^{2}}^{\prime \prime}(1,1)=6, \quad B_{1}=z_{x y}^{\prime \prime}(1,1)=-3, \quad C_{1}=z_{y^{2}}^{\prime \prime}(1,1)=6 . \\
A_{1} C_{1}-B_{1}^{2}=36-9=27>0 \quad \text { and } \quad A_{1}=6>0 .
\end{gathered}
$$

At the point $(1,1)$, we have a minimum and $z_{\min }=z(1,1)=-1$.

For $M_{2}(0,0)$ we have $A_{2}=0, B_{2}=-3, C_{2}=0, A_{2} C_{2}-B_{2}^{2}=-9$. At the point $(0,0)$, the function $z$ does not have an extremum.