Example 1. Investigate the function for extremum

$$
f(x, y)=x^{3}+y^{3}+9 x y
$$
Solution. We find the first and second partial derivatives:

$$
\begin{gathered}
f_{x}^{\prime}(x, y)=3 x^{2}+9 y ; f_{y}^{\prime}(x, y)=3 y^{2}+9 x \\
f_{x x}^{\prime \prime}(x, y)=6 x ; f_{x y}^{\prime \prime}(x, y)=9 ; f_{y y}^{\prime \prime}(x, y)=6 y
\end{gathered}
$$

Setting the first derivatives to zero, we obtain a system of equations to determine the critical points:

$$
\left.\left.\begin{array}{c}
3 x^{2}+9 y=0 \\
3 y^{2}+9 x=0
\end{array}\right\} \text { or } \begin{array}{r}
x^{2}+3 y=0 \\
y^{2}+3 x=0
\end{array}\right\}
$$

Solving for $y$ from the first equation and substituting its expression

$$
y=-\frac{1}{3} x^{2}
$$

into the second equation, we get

$$
\left(-\frac{1}{3} x^{2}\right)^{2}+3 x=0, x^{4}+27 x=0
$$

or

$$
x\left(x^{3}+27\right)=0
$$

from which

$$
x_{1}=0, x_{2}=-3
$$

(We do not consider the complex roots of the equation $x^{3}+27=0$ or $(x+3)\left(x^{2}-3 x+\right.$ $+9)=0$.)

We find the values of $y$ corresponding to the values $x_{1}=0, x_{2}=-3$. From equation (A), we have $y_{1}=0, y_{2}=-3$. We have obtained two critical points $M_{1}(0,0), M_{2}(-3,-3)$.

We compute the values of the second partial derivatives at these points:

$$
\begin{gathered}
A_{1}=f_{x x}^{\prime \prime}(0,0)=0, B_{1}=f_{x y}^{\prime \prime}(0,0)=9, C_{1}=f_{y y}^{\prime \prime}(0,0)=0 \\
A_{2}=f_{x x}^{\prime \prime}(-3,-3)=-18, B_{2}=f_{x y}^{\prime \prime}(-3,-3)=9 \\
C_{2}=f_{y y}^{\prime \prime}(-3,-3)=-18
\end{gathered}
$$

We find the determinant (8.9):

$$
\begin{gathered}
\Delta_{1}=A_{1} C_{1}-B_{1}^{2}=0 \cdot 0-9^{2}=-81 \\
\Delta_{2}=A_{2} C_{2}-B_{2}^{2}=(-18)(-18)-9^{2}=243
\end{gathered}
$$

By the sufficient conditions, we conclude that there is no extremum at point $M_{1}$, since $\Delta_{1}<0$, and there is a maximum at point $M_{2}$, since $\Delta_{2}>0$ and $A_{2}<0$, and

$$
\max f(x, y)=f(-3,-3)=27
$$