Take the version of the Cauchy matrix defined such that $a_{ij} = \dfrac 1 {x_i + y_j}$. Subtract column $1$ from each of columns $2$ to $n$. Thus: {{begin-eqn}} {{eqn | l = a_{ij}       | o = \gets       | r = \frac 1 {x_i + y_j} - \frac 1 {x_i + y_1}       | c =  }} {{eqn | r = \frac {\left({x_i + y_1}\right) - \left({x_i + y_j}\right)} {\left({x_i + y_j}\right) \left({x_i + y_1}\right)}       | c =  }} {{eqn | r = \left({\frac {y_1 - y_j}{x_i + y_1} }\right) \left({\frac 1 {x_i + y_j} }\right)       | c =  }} {{end-eqn}} From Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant. Now: :$1$: extract the factor $\dfrac 1 {x_i + y_1}$ from each row $1 \le i \le n$ :$2$: extract the factor $y_1 - y_j$ from each column $2 \le j \le n$. Thus from Determinant with Row Multiplied by Constant we have the following: :$\displaystyle D_n = \left({\prod_{i \mathop = 1}^n \frac 1 {x_i + y_1}}\right) \left({\prod_{j \mathop = 2}^n y_1 - y_j}\right) \begin{vmatrix} 1 & \dfrac 1 {x_1 + y_2} & \dfrac 1 {x_1 + y_3} & \cdots & \dfrac 1 {x_1 + y_n} \\ 1 & \dfrac 1 {x_2 + y_2} & \dfrac 1 {x_2 + y_3} & \cdots & \dfrac 1 {x_2 + y_n} \\ 1 & \dfrac 1 {x_3 + y_2} & \dfrac 1 {x_3 + y_3} & \cdots & \dfrac 1 {x_3 + y_n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & \dfrac 1 {x_n + y_2} & \dfrac 1 {x_n + y_3} & \cdots & \dfrac 1 {x_n + y_n} \\ \end{vmatrix}$ Now subtract row $1$ from each of rows $2$ to $n$. Column $1$ will go to $0$ for all but the first row. Columns $2$ to $n$ will become: {{begin-eqn}} {{eqn | l = a_{ij}       | o = \gets       | r = \frac 1 {x_i + y_j} - \frac 1 {x_1 + y_j}       | c =  }} {{eqn | r = \frac {\left({x_1 + y_j}\right) - \left({x_i + y_j}\right)} {\left({x_i + y_j}\right) \left({x_1 + y_j}\right)}       | c =  }} {{eqn | r = \left({\frac {x_1 - x_i} {x_1 + y_j} }\right) \left({\frac 1 {x_i + y_j} }\right)       | c =  }} {{end-eqn}} From Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant. Now: :$1$: extract the factor $x_1 - x_i$ from each row $2 \le i \le n$ :$2$: extract the factor $\dfrac 1 {x_1 + y_j}$ from each column $2 \le j \le n$. Thus from Determinant with Row Multiplied by Constant we have the following: :$\displaystyle D_n = \left({\prod_{i \mathop = 1}^n \frac 1 {x_i + y_1}}\right) \left({\prod_{j \mathop = 1}^n \frac 1 {x_1 + y_j}}\right) \left({\prod_{i \mathop = 2}^n x_1 - x_i}\right) \left({\prod_{j \mathop = 2}^n y_1 - y_j}\right) \begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & \dfrac 1 {x_2 + y_2} & \dfrac 1 {x_2 + y_3} & \cdots & \dfrac 1 {x_2 + y_n} \\ 0 & \dfrac 1 {x_3 + y_2} & \dfrac 1 {x_3 + y_3} & \cdots & \dfrac 1 {x_3 + y_n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & \dfrac 1 {x_n + y_2} & \dfrac 1 {x_n + y_3} & \cdots & \dfrac 1 {x_n + y_n} \\ \end{vmatrix}$ From Determinant with Unit Element in Otherwise Zero Row, and tidying up the products, we get: :$D_n = \frac {\displaystyle \prod_{i \mathop = 2}^n \left({x_i - x_1}\right) \left({y_i - y_1}\right)} {\displaystyle \prod_{1 \mathop \le i, j \mathop \le n} \left({x_i + y_1}\right) \left({x_1 + y_j}\right)} \begin{vmatrix} \dfrac 1 {x_2 + y_2} & \dfrac 1 {x_2 + y_3} & \cdots & \dfrac 1 {x_2 + y_n} \\ \dfrac 1 {x_3 + y_2} & \dfrac 1 {x_3 + y_3} & \cdots & \dfrac 1 {x_3 + y_n} \\ \vdots & \vdots & \ddots & \vdots \\ \dfrac 1 {x_n + y_2} & \dfrac 1 {x_n + y_3} & \cdots & \dfrac 1 {x_n + y_n} \\ \end{vmatrix}$ Repeat the process for the remaining rows and columns $2$ to $n$. The result follows. {{qed}} A similar process obtains the result for the $a_{ij} = \dfrac 1 {x_i - y_j}$ form.