Here is an old question: If $\boldsymbol{s_r}$ denote the sum of the $\boldsymbol{r}$ th powers of the roots of the equation
$$\boldsymbol{x^n+p_1x^{n-1}+\cdots +p_n=0}$$ prove that if the coefficients be expressed in terms of $\boldsymbol{s_r}$ then will $$\boldsymbol{\frac{dp_{r+k}}{ds_r}=-\frac{p_k}{r}}$$ [Brioschi.]
If we just consider the case $n=3$ and attempt to verify it for $\frac{dp_3}{ds_2}$ then we have Newton's identities,
$$-3p_3=s_1p_2+s_2p_1+s_3$$ $$-2p_2=p_1s_1+s_2$$ $$-p_1=s_1$$
Which give
$$\frac{dp_2}{ds_2}=-\frac{1}{2}$$ $$\frac{dp_3}{ds_2}=-\frac{p_1}{2}$$ So this seems to verify the question.
However a different solution is possible, using the chain rule and this does not seem to work,
$$\frac{dp_3}{ds_2}=\frac{dp_3}{dx_1}\frac{dx_1}{ds_2}+\frac{dp_3}{dx_2}\frac{dx_2}{ds_2}+\frac{dp_3}{dx_2}\frac{dx_3}{ds_2}$$ And taking
$$\frac{dx_i}{ds_2}=1/\frac{ds_2}{dx_i}=\frac{1}{2x_i}$$ (this is legitimate right ?) I have the sum
$$\frac{dp_3}{ds_2}=\frac{x_2x_3}{2x_1}+\frac{x_1x_3}{2x_2}+\frac{x_2x_2}{2x_3}$$
and this is not equal to the previous answer.
What is the error in this second method ?
Just because the result is so nice, let's try to prove it.
We have the identity $$(x-x_1)\cdots(x-x_n) = x^n + p_1 x^{n-1} + \cdots + p_n$$ so we can also write $$(1- x_1 x)\cdots (1- x_n x) = 1 + p_1 x + p_2 x^2 + \cdots + p_n x^n$$
Consider the above as (formal) series in $x$ and take the $\log$. We get $$\sum_{i=1}^n \log (1 - x_i x) = \log (1 + p_1 x + \cdots + p_n x^n)$$.
Now recall that $\log (1-u) = -(u + \frac{u^2}{2} + \frac{u^3}{3} +\cdots)$, so we get $$ -(s_1 x + \frac{s_2}{2} x^2 + \frac{s_3}{3} x^3 + \cdots ) = \log ( 1 + p_1 x + \cdots + p_n x^n)$$
Now, let $D$ be any derivation. From the above we
$$-\sum_{k\ge 1} \frac{D(s_k) x^k}{k} \cdot \sum_{i=0}^n p_i x^i = \sum_{i=1}^n D(p_i) x^i$$
Now consider $D = \frac{\partial}{\partial s_r}$ for $1\le r \le n$ and identify the coefficients of $x$. We get the required equalities ( and some more).