Finding points on the surface such that their tangent plane is parallel to the plane $xz$ I have to find the points such that the tangent plane of the surface $x^2+3y^2+4z^2-2xy=16$ are parallel to the plane $xz$.

So I compute $\nabla f(x,y,z)= (2x-2y,6y-2x, 8z)$ and therefore $\nabla f(x,y,z)=(0,1,0)$ where $(0,1,0)$ is the normal vector of the plane $xz$ but I have this equation's system.
\begin{eqnarray}
\left.
\begin{array}{rcl}
     2x-2y & = & 0
  \\ 6y-2x & = & 1
  \\ 8z & = & 0
\end{array}
\right\}
\end{eqnarray}
and I conclude the solution and the point that I found is $(\frac{1}{4}, \frac{1}{4}, 0)$ and this point, give the the plane $y=\frac{1}{4}$ but this point isn't on the surface and this plane isn't tangent to the surface. What's wrong? I'll be grateful if you can give some advice with this.
 A: In order for the tangent plane to be parallel to $xz$, $\nabla f(x,y,z)$ must be collinear to $(0,1,0)$, i.e. for some $k\in\mathbb R$
\begin{eqnarray}
\left\{
\begin{array}{rcl}
     2x-2y & = & 0\cdot k
  \\ 6x-2y & = & 1\cdot k
  \\ 8z & = & 0 \cdot k.
\end{array}
\right.
\end{eqnarray}
Notice that if $k = 0$, then we obtain a singular point at which the tangent plane may not exist.
In addition, the point of tangency must belong to the surface, that is, the equation $$x^2+3y^2+4z^2-2xy=16$$ must be satisfied. Thus, we obtain the system of equations
\begin{eqnarray}
\left\{
\begin{array}{rcl}
     2x-2y & = & 0
  \\ 6x-2y & = &  k
  \\ 8z & = & 0
\\ x^2+3y^2+4z^2-2xy&=&16.
\end{array}
\right.
\end{eqnarray}
From the first three equations $x=y$, $z=0$, $k=4x$. Substituting this into the equation of the surface, we obtain
$$
x^2+3x^2-2x^2=16
$$
$$
2x^2=16
$$
$$
x^2=8.
$$
Hence there are two points where the tangent plane is parallel to the plane $xz$: $(2\sqrt2,2\sqrt2,0)$ and $(-2\sqrt2,-2\sqrt2,0)$. Notice that $k=\pm8\sqrt2 \ne 0$.
