Compute the plane by the three points $(x,y,z)$, $(1,0,2)$, $(0,2,2)$ and check that it is tangent to the surface at $(x,y,z)$.
The normal vector to the plane is given by $N_p=(x-1,y-0,z-2)\times(1-0,0-2,2-2)$.
The normal vector to the surface is given by $N_s=(2x,2y,1)$.
These two vectors are parallel, $N_p\times N_s=0$.
$$-x^2+4xy+y^2-4y-1=0,\\
2+2y^2+4x-2xy-2x^2=0,\\
-y(4x^2+4y^2+4)+2x(x^2+y^2+1)=0.$$
Cancelling the third component gives $x=2y$, then the first reduces to $(y-1)(5y+1)=0$, and the second to $-2(y-1)(5y+1)$.
Solutions: $(2,1,-4)$ and $(-\frac25,-\frac15,\frac45)$.
Best Answer
The equation of the tangent plane to the ellipsoid $x^2 + 2y^2 + 3z^2 = 6$ at the point $(x_0,y_0,z_0)$ can be written as $xx_0+2yy_0+3zz_0=6$. Hence it remains to solve the system $$\begin{cases} x_0^2 + 2y_0^2 + 3z_0^2 = 6\\ 6x_0+2\cdot 0y_0+3\cdot 0z_0=6\\ 0x_0+2\cdot 3y_0+3\cdot 0z_0=6 \end{cases}.$$ Can you take it from here?