An easier evaluation of $detlimits_{1leqslant i,jleqslant n}left{frac{x_i-x_j}{x_i+x_j}right}$

To evaluate $D_n(bar{x},bar{y},bar{z})$, we write $$frac{x_i+z_j}{x_i+y_j}=(z_j-y_j)left(frac{1}{x_i+y_j}-frac{1}{y_j-z_j}right),$$ so that $$D_n(bar{x},bar{y},bar{z})=prod_j(z_j-y_j) begin{vmatrix} 1&frac{1}{y_1-z_1}&ldots&frac{1}{y_n-z_n} \1&frac{1}{x_1+y_1}&ldots&frac{1}{x_1+y_n} \cdot&cdots&ddots&cdots \1&frac{1}{x_n+y_1}&ldots&frac{1}{x_n+y_n} end{vmatrix}.$$

Expanding the determinant along the first row, we get $$D_n(bar{x},bar{y},bar{z})=prod_j(z_j-y_j)left(C_n(bar{x},bar{y})+sum_{k=1}^{n}frac{C_n^{(k)}(bar{x},bar{y})}{y_k-z_k}right),$$ where $C_n$ is the well-known Cauchy determinant: $$C_n(bar{x},bar{y}):=det_{1leqslant i,jleqslant n}left{frac{1}{x_i+y_j}right}=frac{prod_{i<j}(x_i-x_j)(y_i-y_j)}{prod_{i,j}(x_i+y_j)},$$ and $C_n^{(k)}$ is obtained by replacing the $k$-th column of $C_n$ by a column of $color{blue}{1}$s: $$C_n^{(k)}(bar{x},bar{y})=lim_{y_ktoinfty}y_k C_n(bar{x},bar{y})=C_n(bar{x},bar{y})frac{prod_i(x_i+y_k)}{prod_{ineq k}(y_k-y_i)}.$$

Thus, we obtain Duparc's result $$D_n(bar{x},bar{y},bar{z})=C_n(bar{x},bar{y})prod_j(z_j-y_j)left(1-sum_ifrac{y_i+x_i}{y_i-z_i}prod_{jneq i}frac{y_i+x_j}{y_i-y_j}right).$$

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Since $R_n(bar{x})=D_n(bar{x},bar{x},-bar{x})$, we're left to show $$sum_iprod_{jneq i}frac{x_i+x_j}{x_i-x_j}=begin{cases}0,&ntext{ is even}\1,&ntext{ is odd}end{cases}.tag{*}label{essential}$$

For this, we do partial fraction expansion of $$F(x):=prod_jfrac{x+x_j}{x-x_j}=A_0+sum_ifrac{A_i}{x-x_i},$$ with $A_0=1$ and $A_i=limlimits_{xto x_i}(x-x_i)F(x)=ldots$ resulting in $$prod_jfrac{x+x_j}{x-x_j}=1+sum_ifrac{2x_i}{x-x_i}prod_{jneq i}frac{x_i+x_j}{x_i-x_j}.$$ To obtain eqref{essential}, it just remains to put $x=0$.