Your answer is correct, but the logic is wrong. You should consider the above surface as a level surface of the function:
$$ f(x, y, z) = x^2 + 2y^2 + z^2 - 2x - 2z - 2$$
Then the gradient should be a vector with 3 components.
$$ \nabla f(x, y, z) = (2x - 2, 4y, 2z - 2) $$
If the tangent plane is horizontal, the gradient must point in the $z$-direction, therefore the $x$ and $y$ components are $0$. So it follows that $x = 1$ and $y = 0$
Finally, the equations are just $z = 3$ and $z = -1$ since horizontal planes have the same $z$ coordinates everywhere.
Another way is to treat $z$ as a function of $x$ and $y$, then set $\partial z / \partial x$ and $\partial z / \partial y$ equal to $0$. However, those expressions are not the same as what you wrote, since they have to be found through implicit differentiation. I'm not sure which method you were trying to use.
Putting $x=y=z=0$ into the equation of the tangent plane, we have
\begin{array}
$f'\left(\frac{x_0}{y_0}\right)(x-x_0) + \left[f\left(\frac{x_0}{y_0}\right)-\frac{x_0}{y_0}f'\left(\frac{x_0}{y_0}\right)\right](y-y_0) - (z-z_0)
&=& f'\left(\frac{x_0}{y_0}\right)(-x_0) + \left[f\left(\frac{x_0}{y_0}\right)-\frac{x_0}{y_0}f'\left(\frac{x_0}{y_0}\right)\right](-y_0) - (-z_0)\\
&=&-x_0f'\left(\frac{x_0}{y_0}\right) -y_0f\left(\frac{x_0}{y_0}\right)+x_0f'\left(\frac{x_0}{y_0}\right) +z_0\\
&=& z_0-y_0f\left(\frac{x_0}{y_0}\right)
\end{array}
But from the given surface $z_0=y_0f\big(\frac{x_0}{y_0}\big)$ and we are done.
Best Answer
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 $(-\frac25,-\frac15,\frac45)$.