Show that the following power series satisfies this functional equation $fleft(frac{2x}{1+x^2}right)=(1+x^2),f(x)$.

Here $$f(x)=frac{tanh^{-1}{x}}{x}$$ $$frac{f[2x/(1+x^2)]}{f(x)}=frac{tanh^{-1}(2x/(1+x^2))}{2x/(1+x^2)}frac{x}{tanh^{-1} x}=(1+x^2).$$ Next the MacLaurin series for $$f(x)=frac{tanh^{-1}x}{x}=sum_{k=0}^{infty} frac{x^{2k}}{(2k+1)!}$$

This is a proof that the identity holds in $mathbb{K}[![x]!]$ for an arbitrary base field $mathbb{K}$ of characteristic $0$, where it is not possible to write $$f(x)=dfrac{1}{color{red}2x},lnleft(dfrac{1+x}{1-x}right)$$ (although we can technically define $ln(1+x)$, $ln(1-x)$, and $lnleft(dfrac{1+x}{1-x}right)$ as power series in $mathbb{K}[![x]!]$). Note that in my comment under the question, I forgot a factor $2$.

Since $f(x)=displaystylesum_{k=0}^infty,frac{x^{2k}}{2k+1}$, we get $$g(x):=frac{1}{1+x^2},fleft(frac{2x}{1+x^2}right)=frac{1}{1+x^2},sum_{k=0}^{infty},frac{1}{(2k+1)},left(frac{2x}{1+x^2}right)^{2k},.$$ Therefore, $$g(x)=sum_{k=0}^infty,frac{2^{2k},x^{2k}}{2k+1},(1+x^2)^{-2k-1}=sum_{k=0}^infty,frac{2^{2k},x^{2k}}{2k+1},sum_{r=0}^infty,binom{-2k-1}{r},x^{2r},.$$ Since $displaystylebinom{-m}{n}=(-1)^n,binom{m+n-1}{n}$, we get $$g(x)=sum_{k=0}^infty,frac{2^{2k},x^{2k}}{2k+1},sum_{r=0}^infty,(-1)^r,binom{2k+r}{r},x^{2r},.$$ That is, $$g(x)=sum_{k=0}^infty,sum_{r=0}^infty,frac{(-1)^r,2^{2k}}{2k+1},binom{2k+r}{r},x^{2(k+r)},.$$ Let $s:=k+r$. Then, $$g(x)=sum_{s=0}^infty,x^{2s},sum_{k=0}^s,frac{(-1)^{s-k},2^{2k}}{2k+1},binom{s+k}{s-k},.$$ In order to prove that $g(x)=f(x)$, we need to show that $$frac{1}{2s+1}=sum_{k=0}^s,frac{(-1)^{s-k},2^{2k}}{2k+1},binom{s+k}{s-k}tag{*}$$ for all $s=0,1,2,ldots$.

However, we are in luck. The equation (*) is an equality for rational numbers, which is the prime field of $mathbb{K}$. Therefore, we can simply prove (*) by using the result when $mathbb{K}=mathbb{R}$. Note that $f(x)=dfrac{1}{2x}lnleft(dfrac{1+x}{1-x}right)$ for $xinmathbb{R}$ such that $0<|x|<1$. Since $$fleft(frac{2x}{1+x^2}right)=frac{1+x^2}{4x},lnleft(frac{1+frac{2x}{1+x^2}}{1-frac{2x}{1+x^2}}right)=frac{1+x^2}{4x},lnleft(frac{1+2x+x^2}{1-2x+x^2}right),,$$ therefore $$begin{align}fleft(frac{2x}{1+x^2}right)&=(1+x^2),left(frac{1}{4x},lnleft(left(frac{1+x}{1-x}right)^2right)right) \&=(1+x^2),left(frac{1}{4x}cdot 2lnleft(frac{1+x}{1-x}right)right) \&=(1+x^2),left(frac{1}{2x},lnleft(frac{1+x}{1-x}right)right) \&=(1+x^2),f(x),.end{align}$$ Thus, (*) holds in $mathbb{R}$, whence (*) is an equality of rational numbers. Consequently, (*) is true in any field $mathbb{K}$ of characteristic $0$. Therefore, the identity $$fleft(frac{2x}{1+x^2}right)=(1+x^2),f(x)$$ holds in $mathbb{K}[![x]!]$ for any field $mathbb{K}$ of characteristic $0$.

Remark. I think there has to be a direct combinatorial or algebraic way to prove (*). My proof of (*) is in a very roundabout manner.

Observe that $g(x)=xf(x)=x+frac {x^3}3+...$ and so for $| x|<1$ you have $g’(x)=1+x^2+x^4+...=frac 1{1-x^2}$. From this you obtain $xf(x)=g(x)=int_0^x frac 1{1-t^2}dt =frac 12 log frac {1+x}{|1-x|}$ and finally $f(x)= frac 1{2x}log frac {1+x}{|1-x|}$ for all $x$ in a nearly of $0$. Can you reach thesis from this?