There is one practice question in which I have to find the points on a graph z = 3x^2 - 4y^2 at which the vector n = <3,2,2> is the normal vector to the tangent plane. Here is how far I got:
z = 3x^2-4y^2 n = <3,2,2> gradient of w = <6x, -8y, -1>
0 = 3x^2-4y^2-z = w(x,y,z)
6x = 3 -8y = 2 -1 = 2
The answer is supposedly -1/4, 1/8, 1/8
You are looking for a point where the gradient of $w$ is equal to $\vec n$. That's too restrictive. You have to look for a point where the gradient of $w$ is parallel to $\vec n$. After all, if $\vec n$ is a normal vector at some point, then $2\vec n$ is also a normal vector at the same point.
In other words, you need to introduce a new variable $k$ and solve $$ \cases{6x = 3k\\-8y = 2k\\-1 = 2k} $$ Once you've found $x$ and $y$ from this equation, you can insert this into the original expression for the graph to find which $z$ coordinate gives you a point on the surface.