Jacobian of parametrized ellipsoid with respect to parametrized sphere I'm not even sure how best to phrase this question, but here goes. Given $\theta$ (elevation) and $\phi$ (azimuth), the unit sphere can be parametrized as 
$
x = \cos(\theta)\sin(\phi) \\
y = \cos(\theta)\cos(\phi) \\
z = \sin(\theta).
$.
A general ellipsoid can then be written as $X = ax$, $Y = by$, $z = cz$.
I'm trying to find the Jacobian that tells you how the sphere was transformed to the ellipsoid. In my mind, this involved computing the following matrix
$
\frac{\partial X}{\partial x},\frac{\partial X}{\partial y}, \frac{\partial X}{\partial z} \\ \frac{\partial Y}{\partial x}, \frac{\partial Y}{\partial y}, \frac{\partial Y}{\partial z} \\
\frac{\partial Z}{\partial x},\frac{\partial Z}{\partial y}, \frac{\partial Z}{\partial z}
$


*

*Is this correct, or should the "matrix" only have the diagonal entries?

*If it is not correct, would this idea be correct for an implicit surface?

*If it is correct, how do I do the differentiation for the off-diagonal entries? Could you work out a single example for differentiating a function with respect to $y$ and $z$?
I've been cracking my head on this one, though it seems like it should be extremely simple.

The map $(x,y,z)\mapsto (ax,by,cz) = (X,Y,Z)$ takes three variables to three variables, rather than the two variables of your parametrization. The Jacobian matrix of this three-dimensional transformation is 
$$J = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}.$$
(So the answer to your first question is, "both." The off-diagonal entries are $0$.)
What maybe is confusing you is that this is a three-dimensional transformation and makes no reference to $\theta$ or $\phi$. (Is this why you expect there to be off-diagonal entries?) To see what's going on with the ellipsoid, in particular, you need to find the Jacobian of your parametrization (which tells you how the $\theta$ and $\phi$ directions are distorted by embedding them in three-space) and then compose it with the matrix $J$. This will be equivalent to differentiating the composition $(X(\theta,\phi), Y(\theta,\phi), Z(\theta,\phi))$.

Just for kicks, here's another attack that doesn't make any explicit use of calculus. Since the unit sphere is defined implicitly by $x^2 + y^2 + z^2 = 1$, the tangent space to the sphere at the point $p$, is all the vectors perpendicular to $(x,y,z)$. Multiply these vectors by $J$ and you have the vectors to the ellipsoid at $(X,Y,Z)$.
