Describe the curves followed by the raindrops in the following question Problem: It is raining and rainwater is running off an ellipsoidal
dome with equation $4x^2 + y^2 + 4z^2 = 16$, where
z ≥ 0. 
Given that gravity will cause the raindrops to
slide down the dome as rapidly as possible, describe
the curves whose paths the raindrops must follow.
(Hint: You will need to solve a simple differential
equation.)
Here's how I approached this problem:
a) Let $F= 4x^2 + y^2 + 4z^2$ 
and $(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z})$ = $(8x, 2y, 8z)$
b) Also, the path followed by the raindrops should be $(\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y})$
I am stuck now. I can't think of a way to relate the two things that I have. 
Update:
I considered the surface $S := [X\in R^3| F(X) = 16] $ 
also, I set X = (x, y, z(x,y))
Then $DF(X) = D(16)= 0$
Thus, $(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z})$ 
 $\begin{pmatrix}
   \frac{\partial x}{\partial x}& \frac{\partial x}{\partial y}\\
   \frac{\partial y}{\partial x} &\frac{\partial y}{\partial y}\\
   \frac{\partial z}{\partial x}& \frac{\partial z}{\partial y}\\
   \end{pmatrix}$ = 0
OR, $(8x, 2y, 8z)\begin{pmatrix}
   1& 0\\
  0 &1\\
   \frac{\partial z}{\partial x}& \frac{\partial z}{\partial y}\\
   \end{pmatrix}$ = 0
This gives me the following
$\frac{\partial z}{\partial x}= \frac{-x}{z} $ and $\frac{\partial z}{\partial y} = \frac{-y}{4z}$
Do i solve these two differential equations to get the curves that i want?
I if i divide these two then, I can also get $\frac{\partial y}{\partial x}=$$\frac{4x}{y}$ so do i solve this ODE to get my curves? I am not sure.
 A: I believe when $z<0$ it would drop. So we will assume $z>0$. On the path the tangent vector should be 
$$\frac{dy}{dx}=\frac{z_y}{z_x}$$
where 
$$(z_x,z_y)=(\frac{-x}{\sqrt{\frac{16-4x^2-y^2}{4}}},\frac{-\frac{1}{4}y}{\sqrt{\frac{16-4x^2-y^2}{4}}})$$
Hence 
$$\frac{dy}{dx}=\frac{\frac{1}{4}y}{x}$$
You can then solve this ODE to get the curve on the surface.
Edit: Your edit is almost correct, but you confused the tangent vector with the path vector. $(z_x,z_y)$ is the direction, so it is the tangent vector of the path. Now if the raindrop follows $(x(t),y(t))$, then $(x'(t),y'(t))=(z_x,z_y)$. That's how I got
$$\frac{dy}{dx}=\frac{z_y}{z_x}$$
You had the reciprocal instead.
A: Your condition $D({\bf X})=0$ just ensures that the raindrops stay on the ellipsoid $S$. But in reality they think in which direction they should go: It is the direction of fastest descent. Solving $F(x,y,z)=16$ for $z$ gives
$$z=f(x,y):=\sqrt{4-x^2-{y^2\over4}}\ .$$
The $(x,y)$-direction of fastest $z$-descent on $S$ is given by the negative gradient
$$-\nabla f(x,y)=\left({2x\over 2f(x,y)}, \>{2y/4\over 2f(x,y)}\right)={1\over 4f(x,y)}\>(4x,y)\ .$$ The slope ($y$ versus $x$) of this neggradient at $(x,y)$ is
$$y'={y\over 4x}\ .$$
This is the ODE we were looking after. Its solutions are the curves
$$y(x)=C\>|x|^{1/4}, \qquad C\in{\mathbb R}\ .$$
