Given $a\in\mathbb{R}^{mn\times n}$, find a $C\in\mathbb{R}^{n}$, $x\in\mathbb{R}^{m\times n}$ such that $$ 0 = f_{k}(\boldsymbol{C}, \boldsymbol{x}):=\sum_{i=1}^{m} C_{i} \left(\prod_{j=1}^{n} a_{kj}^{x_{i,j}} - 1\right) $$ for all $k\in\{1,\dots,mn\}$.
(The rows of $a_{kj}$ are vectors of solutions to a system of differential equations, and the goal of finding the roots of the above system is to find invariant quantities of those diff. eq.)
To show that the system has a singular Jacobian, I have a test setup where $a_{k1} = (1-2t_{k})^{-0.5}$, $a_{k2} = (1-2t_{k})^{-1}$ and the rest zero (only $i=1$, but it is still singular if you include the sum). It is in this case easy to check that the Jacobian is singular irregardless of the values of $\boldsymbol{C}$ and $\boldsymbol{x}$. The solution is any $x$ with $2x_{11} - x_{12}=0$ and $C$ anything, and I'm guessing that the Jacobian is singular because this does imply infinite solutions.
I just need to be able to find a couple of them (that are non-trivial such as $C_{i} =0$, and I can go from there. Most root-finding methods I found that can deal with singular Jacobians only mention singularities at the solution, not everywhere. Are there any methods that can tackle this problem? I should also mention that the system should eventually get quite large (order of 1000 equations).