A variation of Cauchy's determinant Prove the following identity:
$$\det_{_{1\leq i,j \leq n}}\left(\frac{1}{(x_i+y_j)^2}\right)=\det_{_{1\leq i,j \leq n}}\left(\frac{1}{x_i+y_j}\right)\text{perm}_{_{1\leq i,j \leq n}}\left(\left(\frac{1}{x_i+y_j}\right)\right)$$
where $\text{perm}(A)$ is the permanent of a matrix $A$ defined as:
$$\text{perm}(A)=\sum_{\sigma\in S_n}\prod_{i=1}^{n}a_{i,\sigma(i)}$$
the sum here extends all elements of the symmetric group $S_n$, i.e. over all permutations of the numbers $1,2,\cdots, n$. 
here is some reference: 

*Permanent

*Symmetric_group
thanks very much
 A: Let $*$ denote the Hadamard product i.e $(A*B)_{ij} := a_{ij} b_{ij}$. For a $\sigma \in S_n$, define $(A_{\sigma})_{ij } := a_{i \sigma(j)}$.Some straightforward computations reveal that
$\displaystyle{\det(A) \text{Perm} (A) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \det(A * A_{\sigma})}$. 
Finally let $(A)_{ij} = (x_i + y_j)^{-1}$. Notice that the given problem follows from the following claim. 
Claim: If $\sigma \in S_n$ is not the identity permutation, $\det(A*A_{\sigma}) = 0$. 
Proof: Let $\sigma \ne \text{id}$, contain the cycle $(u_1,\ldots,u_k) $. We show that the space spanned by the columns $ \left \{ (A*A_{\sigma})_{u_i} \right \}$ has dimension at most $k-1$, which will imply that $\det(A*A_{\sigma}) = 0$. Note that it suffices to consider the case when $u_i = i, \forall i \in (1,\ldots, k)$. Form a matrix with those columns and multiply the $i-$th row by $\prod_{t=1}^k (x_i + y_t)$. Now we get the matrix $M$ such that $\displaystyle{M_{ij} = \prod_{t \in [k]\setminus {(j,j+1)}} (x_i + y_t)}$ which has same rank as the previous matrix. 
Note that $\forall i \in (1, \ldots, n)$ we have ${x_i + y_j = p_j (x_i + y_1) + q_j (x_i + y_2)} $ where $\displaystyle{}p_j = \frac{y_j -  y_2}{y_1 -y_2}$ and $\displaystyle{q_j = \frac{y_1 - y_j}{y_1 -y_2}}$ for any fixed $j \in (1, \ldots, k)$
So $ M_{ij} = \prod_{t \in [k]\setminus {(j,j+1)}} (x_i + y_t)$
$ = \prod_{t \in [k]\setminus {(j,j+1)}} (p_t (x_i + y_1) + q_t (x_i + y_2)) = \sum_{r=0}^{k-2} C_{m,j}  (x_i + y_1)^r (x_i + y_2)^{k-2 - r}$
This shows that $\text{rank} (M) \le k-1$ proving the claim.
