Example 1. Find the direction vector of the normal to the ellipsoid $x^{2}+2 y^{2}+3 z^{2}=6$ at the point $M_{0}(1,-1,1)$.
Solution. First of all, the point $M_{0}$ lies on the ellipsoid, which can be verified by substituting its coordinates into the given equation. This equation can be rewritten as

$$
x^{2}+2 y^{2}+3 z^{2}-6=0
$$

By comparing this equation with equation (8.1), we conclude that

$$
F(x, y, z)=x^{2}+2 y^{2}+3 z^{2}-6
$$

We find the partial derivatives of the function (A):

$$
\frac{\partial F}{\partial x}=2 x, \frac{\partial F}{\partial y}=4 y, \frac{\partial F}{\partial z}=6 z
$$

and calculate their values at the point $M(1,-1,1)$, i.e., when $x_{0}=1, y_{0}=$ $=-1, z_{0}=1:$

$$
\left(\frac{\partial F}{\partial x}\right)_{0}=2,\left(\frac{\partial F}{\partial y}\right)_{0}=-4,\left(\frac{\partial F}{\partial z}\right)_{0}=6
$$

Therefore, the direction vector of the normal is

$$
\bar{m}=\{2,-4,6\} \text {. }
$$

Remark. The vector $\overline{m^{\prime}}=\{1,-2,3\}$ or any other vector collinear with the vector $\bar{m}=$ $=\{2,-4,6\}$ can also be taken as the direction vector of the normal.