Looking at old things in new ways!

Firstly,

$\large \displaystyle 1+x+x^{2}+x^{3}+... \infty = \frac{1}{1-x} ; -1<x<1$

diffrentiating both sides wrt $\large \displaystyle x$ , twice,

$\large \displaystyle \frac{2!}{(1-x)^{3}} = f(x)$

$\large \displaystyle f^{'}(x) = \frac{3!}{(1-x)^{4}}$

$\large \displaystyle \frac{d^{r}f(x)}{dx^{r}}=\frac{(r+2)!}{(1-x)^{r+3}}$

it is evident that the sum $\large \displaystyle \frac{1}{f(x)}+\frac{1}{f^{'}(x)}+\frac{1}{f^{''}(x)}+\frac{1}{f^{'''}(x)}+... \infty$ is Taylor's series expansion for $\large \displaystyle (1-x)[e^{1-x}-(1-x)-1]$

$\large \displaystyle g(\frac{1}{3}) = \frac{2}{9}[3e^{\frac{2}{3}}-5]$

$\large f(x)=2+6x+12x^{2}+20x^{3}+30x^{4}+...........$ , doing some algebraic manipulation, $$ first multiplying by x then 1-x , then x(1-x) st different times in $f(x)$ and then subtracting and adding $$ , we get that $\large f(x)$ = $\large \displaystyle \frac{2}{(1-x)^{3}}$ $$ and then differentiate as @Rohith M.Athreya has procedded !