How to find the direction vector of a ball falling off an ellipsoid? 
A tiny ball is placed in top of an ellipsoid $3x^2+2y^2+z^2=9$ at $(1,1,2)$. Find the three-dimensional vector $\underline u$ in whose direction the ball will start moving after the ball is released.

I feel this problem involves usage of gradients but not sure how to tackle it.
EDIT the solution shouldn't use physics knowledge and has to be based on directional derivatives and/or gradients.
EDIT 1 I've finally come up with the "no physics solution" however it is different from the accepted answer, I'd appreciate if other members confirm if the accepted answer is correct.
One potential flaw with the accepted answer is that it's not using the $9$ from the original equation $3x^2+2y^2+z^2+\mathbf{9}=0$. 
Anyway this is my take: 
The $xy$ direction in which the ball will fall is $-\nabla f(1,1)$. 
$f_x=-\frac{3x}{\sqrt{9-3x^2-2y^2}}\stackrel{we.plug.in.x=1}{=}-\frac{3}{2}$. 
Similarly, $f_y=-1$ therefore $-\nabla f(1,1)=\langle 3/2,1 \rangle$.
Let the 3d vector we're after be $d=\langle 3/2, 1, a \rangle$. Notice that $d$ is perpendicular to the normal vector of the tangential plane $n=\langle 6x,4x,-1 \rangle=\langle 6,4,-1 \rangle$ so $d\cdot n=0$ therefore $a=13$ so the final result is $d=\langle \frac{3}{2}, 1,13\rangle$. 
 A: Another derivation (besides the answer by Oldboy) is to decompose the gravity vector into the normal (to surface) and parallel (to surface) components.
The normal component is $(\vec{G}\cdot \vec{n})\cdot \vec{n}$, the parallel component is $\vec{G}-(\vec{G}\cdot \vec{n})\cdot \vec{n}$. The parallel direction is the falling direction (same as Oldboy's answer).
A: I will assume that z-axis is oriented vertical upwards.
$f(x,y,z) = 3x^2+2y^2+z^2 - 9 = 0$
Vector $\vec{N}$ perpendicular to the surface $f$ at point $(x,y,z)$ is defined by function gradient:
$\vec{N} = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} + \frac{\partial f}{\partial y} \vec{j} + \frac{\partial f}{\partial z} \vec{k} = 6 x \vec{i} + 4y \vec{j}+2z \vec{k}$ 
At point $(1,1,2)$:
$\vec{N} = 6 \vec{i} + 4 \vec{j} + 4 \vec{k} $
Unit vector corresponding to acceleration of Earth's gravity is:
$\vec{G} = - \vec{k}$
Calculate cross product of vectors $\vec{N}$ and $\vec{G}$. That vector is perpendicular to both vectors and also tangential to the surface:
$\vec{T} = \vec{G} \times \vec{N} = -\vec{k} \times (6 \vec{i} + 4 \vec{j} + 4 \vec{k}) = 4 \vec{i} - 6 \vec{j}$
Calculate vector $\vec{P}$:
$ \vec{P} = \vec{N} \times \vec{T} = (6 \vec{i} + 4 \vec{j} + 4 \vec{k}) \times (4 \vec{i} - 6 \vec{j}) = 24 \vec{i} + 16 \vec{j} - 52 \vec{k}$
This vector is also tangential to the surface $f$ and lies in the plane defined by vectors $\vec{G}$ and $\vec{N}$ and therefore describes the direction of ball motion. You can normalize this vector if necessary.
