A problem from <> by Roytvarf, Birkhauser I got a problem, which turned to be from the book "Thinking in Problems How Mathematicians Find Creative Solutions" by Roytvarf,  Chapter One, Jacobi Identities and Related Combinatorial Formulas :
The problem is asking to prove that
$$
\sum_{i=1}^{n}
\frac{x_i^m}{\prod_{j\neq i}(x_i-x_j)}=0 \ \text{ for }\ m < n-1.
$$
The author gave hint that (I will copy the original words because I have difficulties in following it): "Analytic method: using implicit function theorem : The left-hand sides are proportional to the first derivatives of the functions $\sum_{i=1}^n x_i^{m+1}$, with respect to the constant term of the polynomial $F(z):={\prod_i (z-x_{i})}$, while the other coefficients are kept fixed "
What does it mean ? The constant term of $\prod_i (z-x_i)$ is $\prod_i x_i$, so I try to take the partial derivative as
$$
\frac{\partial \sum_i x_i^{m+1}}{\partial \prod_i x_i}
=\sum_j \left( \frac{\partial \sum_i x_i^{m+1}}{\partial x_j}
\frac{\partial x_j}{\partial \prod_i x_i} \right)
=\sum_j \left( \frac{\partial(m+1)x_j^m}{(\prod_i x_i)/x_j} \right)
=\sum_j \left( \frac{(m+1)x_j^{m+1}}{(\prod_i x_i)} \right).
$$
However it doesn't seem to contain the factor of left handside of the equation.  Also I am confused about the meaning that "while the other coefficients are
kept fixed"
I think most of all I just don't see how implicit function theorem is related to the hint here.
Can someone help me to understand? Thanks a lot!
 A: This solution is similar to what you are saying, but not phrased in your language of implicit function theorem.
Consider the polynomial
$$ \sum_{i=1}^n \left[ x_i ^m  \prod_{j < k, j\neq i, k\neq i} (x_j-x_k)\right] $$
If $x_i = x_j$, then you should verify that the polynomial is equal to 0 (many terms with $x_i-x_j)$ are zero, and then ensure that the two remaining terms exactly cancel out). Hence, $(x_i-x_j)$ is a factor of the above polynomial.
This is true for all ${n\choose 2}$ combinations of $i,j$, and thus we know ${n\choose 2}$ factors of the polynomial. However, since $m < n-1$, the degree of this polynomial is strictly less than ${n-1 \choose 2} + n-1 = { n \choose 2}$. Hence, it must be the zero polynomial.
Now, divide the polynomial by $\prod_{j\neq k} (x_j-x_k)$ and we are done.
A: You consider the map $c\colon (x_1,\ldots,x_n)\mapsto (a_0,\ldots,a_{n-1})$ where $x^n+a_{n-1}x^{n-1}+\ldots+a_0=(x-x_1)\cdots(x-x_n)$.
For any point $\mathbf x=(x_1,\ldots,x_n)$ with all entries different(!), you can find an inverse map $c^{-1}$ from a neighbourhood of $c(\mathbf x)$ to a neighbourhood of $\mathbf x$. Of course you can combine this with $s_m\colon(x_1,\ldots,x_n)\mapsto \sum_i x_i^{m+1}$ and then consider $\frac \partial {\partial a_0}(s_m\circ c^{-1})$
