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.
$$\mathrm{f}_x = \mathrm{f}_y = 0$$
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.
Best Answer
The surface can be expressed as $y - x^2 - 3z^2 = 0$, and the gradient of $y - x^2 - 3z^2$ will be perpendicular to the surface at $(x,y,z)$. Note that $(-2x,1,-6z)$ is this gradient.
In order for a coordinate hyperplane to be tangent to the surface at $(x,y,z)$, this gradient vector also has to have two entries that are zero. The only way this can happen is if $x = z = 0$, which corresponds to gradient vector $(0,1,0)$. And indeed when $x = z = 0$, one has $y = 0$ and the tangent plane to the surface can be verified through the usual formula to be the coordinate plane $y = 0$.