= Solution
A tangent displacement has the form $(dx,dy,z_xdx+z_ydy)$. At the specified point, $z_y=0$ and $z_x=q>0$, so the fastest decrease of height per unit surface distance has $dy=0$ and $dx<0$. The unit downhill tangent is $(-1,0,-q)/\sqrt{1+q^2}$. Therefore, for speed $u$ measured along the surface,
$$
\boxed{v=-\frac{u}{\sqrt{1+e^{-1}}}(1,0,e^{-1/2}).}
$$
Its <Euclidean norm> is $u$, and it is orthogonal to the surface normal. The general construction is <steepest descent on a graph surface>; using a horizontal speed equal to $u$ would overestimate the full three-dimensional speed.
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