I think the first thing to iron out is that $\Omega$ should be open, since compact would be closed and then even defining $H^1(\Omega)$ is not trivial. Maybe assume $\Omega \subset \mathbb{R}^N$ is open and $J,f$ being $C^1(\overline{\Omega})$.
However, this is not the issue you are interested in. You want to know the relationship between
$\int_\Omega \nabla u MM^t \nabla u\;dx$
and
$||\nabla u||_{L^2(\Omega) }$.
Given $M$, can you show that the eigenvalues of $MM^t$ are strictly positive? The ellipticity constant is just a statement on the eigenvalues of the matrix associated with the coefficients. So what you are looking to do is to prove that this matrix is strictly positive definite. I hope you can compute this successfully.
EDIT: The result you are looking for in this case is to show that $\xi MM^t\xi \geq c$ for all $\xi \in S^{N-1}$, which would imply the coercivity you are looking for. A priori, $MM^t$ positive definite means you can this inequality with $c=0$, but you need a little better if this is your approach. I imagine For example, if the hypothesis you have make it possible eigenvalues of $MM^t$ are ${\lambda_i}_{i=1..N}$, with $\min_i \lambda_i = \tilde{\lambda}>0$, and these eigenvalues correspond to prove eigenvectors $x_i$ which form an orthogonal basis for $\mathbb{R}^N$, then this slightly stronger (is true, as follows.
Any $x\in S^{N-1}$ can be represented as $x=\sum_{i=1}^N c_i x_i$, where $\sum_i c_i^2=1$, and necessarywe compute $MM^t x = \sum_i c_i MM^t x_i = \sum_i c_i \lambda_i x_i$, from my point of view) resulttherefore $xMM^tx = \sum_i c_i^2 \lambda_i \geq \tilde{\lambda}$, where we have used $\sum_i c_i^2=1$.
This is only an example - you should verify that the eigenvalues correspond to orthonormal eigenvectors to do this, or adapt the proof to something which is close and true in your case.
