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I took $3$ random polynomials with non zero roots one having even degree and two having odd degrees

1. 
   $f(x)=colorred4x^2-(4sqrt3+12)x+12sqrt3$ having roots $colorblue3,sqrt3$ and leading coefficient $colorred4$ and calculated values of $xf'(x)$$(f'(x)$ is the derivative of $f(x))$ at both roots which are $3f'(3)$ and $sqrt3f'(sqrt3)$ and then sum of their reciprocals $frac13f'(3)+frac1sqrt3f'(sqrt3)=frac-112sqrt3=frac-1colorred4left(frac1colorblue3cdotsqrt3right)$ then repeated same thing for


2. 
   

   $g(x)=colorred1x^3-frac203x^2-12x+frac323$ having roots $colorblue8,-2,frac23$

   
   
   

   $frac18g'(8)+frac1-2g'(-2)+frac1frac23g'(frac2 3)=frac1colorred1left(frac1colorblue8cdot-2cdotfrac2 3right)$

   
   


3. 
   

   $h(x)=colorred1x^5+41x^4+137x^3-1601x^2-1818x+3240 $ having roots $colorblue1,-2,5,-9,-36$

   
   
   

   $frac11h'(1)+frac1-2h'(-2)+frac15h'(5)+frac1-9h'(-9)+frac1-36h'(-36)=frac1colorred1left(frac1colorblue1cdot-2 cdot5cdot-9cdot-36right)$

   
   

Is this true for all polynomials? Is there any known result?

This results from partial fraction decomposition.

Suppose $g(x)$ has no repeated roots, and without loss of generality is monic (leading coefficient $1$). Then it factors as $displaystyle g(x)=prod_r(x-r)$ over all roots $r$ and

$$ frac1g(x)=sum_rfracc(r)x-r $$

for some constants $c(r)$, one for each root $r$. To find the constant $c(s)$ for a specific root $s$, first multiply the equation by the factor $(x-s)$,

$$ fracx-sg(x)=c(s)+sum_rne s c(r)fracx-sx-r $$

then "evaluate" i.e. take the limit $xto s$ to obtain

$$ frac1g'(s)=c(s). $$

Therefore we may plug $x=0$ into

$$ frac1g(x) =sum_r frac1g'(r)(x-r) $$

and manage negative signs to get

$$ frac(-1)^deg gdisplaystyle prod r = -sum_r frac1rg'(r). $$

When $deg g$ is odd, all signs can go away.

It is true for all polynomials with non-zero simple roots.

This follows from the barycentric form of Lagrange interpolation:
$$
L(x)=ell (x)sum _j=1^nfrac w_jx-x_jy_j
$$
where
$$
ell (x)=(x-x_1)(x-x_1)cdots (x-x_n),
quad
w_j=frac 1ell '(x_j)
$$
Therefore, for the constant function $1$ evaluated at $x=0$ we have
$$
1= L(0) = -ell (0)sum _j=1^nfrac 1x_jell '(x_j)
$$
For a polynomial $f$ with simple roots $x_1, dots, x_n$, we have
$$
f(x)=a(x-x_1)(x-x_1)cdots (x-x_n)=aell (x)
$$
and so
$$
sum _j=1^nfrac 1x_jf'(x_j)
=sum _j=1^nfrac 1x_jaell '(x_j)
=-frac1aell(0)
=-frac1(-1)^nax_1 cdots x_n
=frac(-1)^n+1ax_1 cdots x_n
$$
This can also be written as
$$
sum _j=1^nfrac 1x_jf'(x_j)
=-frac1f(0)
$$