Show that the following power series satisfies this functional equation $f\left(\frac{2x}{1+x^2}\right)=(1+x^2)\,f(x)$. 
Show that the following power series satisfies this functional equation $$f\left(\dfrac{2x}{1+x^2}\right)=(1+x^2)f(x)\,,$$
where the series given is $$f(x)= 1+\dfrac{1}{3}x^2+\dfrac{1}{5}x^4+\dfrac{1}{7}x^6+ \cdots\,.$$

I can painstakingly get a relation between the derivatives such as $3f^{(2)}(0) = 2f(0)$ but I was hoping for a better approach.
I would appreciate it if someone could give me a hint. I prefer hints to complete solutions.
 A: 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?
A: 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 $$f(x)=\dfrac{1}{\color{red}2x}\,\ln\left(\dfrac{1+x}{1-x}\right)$$ (although we can technically define $\ln(1+x)$, $\ln(1-x)$, and $\ln\left(\dfrac{1+x}{1-x}\right)$ as power series in $\mathbb{K}[\![x]\!]$).  Note that in my comment under the question, I forgot a factor $2$.
Since $f(x)=\displaystyle\sum_{k=0}^\infty\,\frac{x^{2k}}{2k+1}$, we get
$$g(x):=\frac{1}{1+x^2}\,f\left(\frac{2x}{1+x^2}\right)=\frac{1}{1+x^2}\,\sum_{k=0}^{\infty}\,\frac{1}{(2k+1)}\,\left(\frac{2x}{1+x^2}\right)^{2k}\,.$$
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 $\displaystyle\binom{-m}{n}=(-1)^n\,\binom{m+n-1}{n}$, 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 $f(x)=\dfrac{1}{2x}\ln\left(\dfrac{1+x}{1-x}\right)$ for $x\in\mathbb{R}$ such that $0<|x|<1$.  Since
$$f\left(\frac{2x}{1+x^2}\right)=\frac{1+x^2}{4x}\,\ln\left(\frac{1+\frac{2x}{1+x^2}}{1-\frac{2x}{1+x^2}}\right)=\frac{1+x^2}{4x}\,\ln\left(\frac{1+2x+x^2}{1-2x+x^2}\right)\,,$$
therefore
$$\begin{align}f\left(\frac{2x}{1+x^2}\right)&=(1+x^2)\,\left(\frac{1}{4x}\,\ln\left(\left(\frac{1+x}{1-x}\right)^2\right)\right)
\\&=(1+x^2)\,\left(\frac{1}{4x}\cdot 2\ln\left(\frac{1+x}{1-x}\right)\right)
\\&=(1+x^2)\,\left(\frac{1}{2x}\,\ln\left(\frac{1+x}{1-x}\right)\right)
\\&=(1+x^2)\,f(x)\,.\end{align}$$
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
$$f\left(\frac{2x}{1+x^2}\right)=(1+x^2)\,f(x)$$
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.
A: 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 $$f(x)=\frac{\tanh^{-1}x}{x}=\sum_{k=0}^{\infty} \frac{x^{2k}}{(2k+1)!}$$
