As you know that:
$$ \arctan(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} \quad\Rightarrow\quad \arctan (2x) = \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n+1}}{2n+1} $$
Also, you have:
$$ \frac{d}{dx}\left[\arctan(x)\right] = \frac{1}{1+x^2} \quad\Rightarrow\quad \frac{d}{dx}\left[\arctan(2x)\right] = \frac{2}{1+4x^2} $$
Thus,
$$
\begin{align}
\color{red}{\frac{1}{1+4x^2}} &= \frac{1}{2}\,\frac{d}{dx} \left[\color{white}{\frac{}{}}\arctan(2x)\color{white}{\frac{}{}}\right] = \frac{1}{2}\,\frac{d}{dx} \left[\,\sum_{n=0}^{\infty}(-1)^n\,\frac{(2x)^{2n+1}}{2n+1}\,\right] \\[2mm]
&= \frac{1}{2}\,\sum_{n=0}^{\infty}(-1)^n\,\frac{\left(\,(2x)^{2n+1}\,\right)'}{2n+1} \\[2mm]
&= \frac{1}{2}\,\sum_{n=0}^{\infty}(-1)^n\,\frac{{2\,(2n+1)}\,(2x)^{2n}}{2n+1} \\[2mm]
&= \color{red}{\sum_{n=0}^{\infty}(-1)^n\,(2x)^{2n}}
\end{align}
$$
And with similar analysis, you should get:
$$ \frac{1}{1+9x^2}=\sum_{n=0}^{\infty}(-1)^n\,(3x)^{2n} $$