HINT
Your surface is given by $\mathrm{f}(x,y,z)=0$, where
$$\mathrm{f}(x,y,z) = x^2−7xy−y^2−46x+2y-z$$
The gradient vector $\nabla\mathrm{f} = \left(\mathrm{f}_x,\mathrm{f}_y,\mathrm{f}_z\right)$ is, if non-zero, perpendicular to the the tangent plane. The tangent plane is horizontal if, and only if, it is perpendicular to the $z$-axis. Hence, we need
$$\nabla\mathrm{f} \propto (0,0,1)$$
We need to find the partial derivatives $\mathrm{f}_x$, $\mathrm{f}_y$ and $\mathrm{f}_z$ and check when $\nabla\mathrm{f} \propto (0,0,1)$, i.e.
$$\mathrm{f}_x = \mathrm{f}_y = 0$$
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
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.