For the special case where all $c_{i,j}$'s are equal to 1 and $m=2n-1$, take a look at Ramanujam's paper: http://ramanujan.sirinudi.org/Volumes/published/ram03.pdf. Needless to say, it is an ingenious method. The steps are the following:

(1) The key idea is to recognize that the coefficients (w.r.t $\theta$, upto the $2n-1$ order) of $\sum_{k=1}^n\frac{x_k}{1-\theta y_k}$ in the series expansion would be the LHS of the nonlinear equations.  

(2) Now, $\sum_{k=1}^n\frac{x_k}{1-\theta y_k} = d_1 + d_2\theta + \cdots + d_{2n}\theta^{2n-1} + \cdots = \frac{A_1 + A_2\theta + \cdots + A_{n-1}\theta^{n-1}}{B_1 + B_2\theta + \cdots + B_{n-1}\theta^{n-1}}$.

(3) Multiplying by $B_1 + B_2\theta + \cdots + B_{n-1}\theta^{n-1}$ on either side and comparing coefficients, gives linear equations in $A_i$'s and $B_i$'s.

(4) If a solution exists, one can then determine partial fractions (in $\theta$), and the coefficients would be the answers. 

For your case, let me assume (i) $m=2n-1$ and (ii) that the matrix $C\in R^{2n^2,2n^2}$ defined by its $k^{th}$ row as $[\underbrace{0\cdots0}_{n(k-1)}~c_{k,1}\cdots c_{k,n}~0\cdots 0]$ is invertible. Similarly let $I\in R^{2n^2,2n^2}$ be defined by its $k^{th}$ row as $[\underbrace{0\cdots0}_{n(k-1)}~1\cdots 1~0\cdots 0]$. Let $z = [x_1\cdots x_n~\cdots ~x_1y_1^{2n-1}\cdots x_ny_n^{2n-1}]^\top$. Then the set of equations can be written as $Cz = d$. This is equivalent to $z = C^{-1}d$, and thus $Iz = IC^{-1}d$. This puts your system in Ramanujam's format.

All this is only a suggestion for the most general case. Hope it helps.