We have
$$\frac1{(1-t)^2}=\sum_{m=0}^\infty (m+1)t^{m},\qquad (|t|<1)$$
Put $t=\left(\dfrac{2y}{(1+y)^2}\right)x$, where $0<|y|<1$,
$$\frac1{\left(1-\left(\frac{2y}{(1+y)^2}\right)x\right)^2}=\sum_{m=0}^\infty (m+1)\left(\dfrac{2y}{(1+y)^2}\right)^mx^{m}.\qquad (|x|<\frac{(1+y)^2}{2|y|})$$
Multiply $(1+y)^{2n-1}(1-y)$ on both sides,
$$\frac{(1+y)^{2n+3}(1-y)}{((1+y)^2-2yx)^2}=\sum_{m=0}^\infty(m+1)\big((1+y)^{2n-2m-1}(y^m-y^{m+1})\big)(2x)^m.\qquad (|x|<\frac{(1+y)^2}{2|y|})\tag{*}$$

Consider the coefficient of $y^n$ in $(*)$,

when $m<n$, the coefficient is given by
$$\binom{2n-2m-1}{n-m}y^{n-m}\cdot y^m-\binom{2n-2m-1}{n-m-1}y^{n-m-1}\cdot y^{m+1}=0;$$

when $m=n$,
\begin{align*}
(1+y)^{-1}(y^n-y^{n+1})=\left(\sum_{k=0}^\infty (-y)^k\right)(y^n-y^{n+1})=y^n+\underbrace{\text{something}}_{\text{“deg of $y$” $>n$}},
\end{align*}
so the coefficient is $(n+1) (2x)^n$;

when $m>n$, $2n-2m-1<0$,
\begin{align*}
(1+y)^{2n-2m-1}(y^m-y^{m+1})=\left(\sum_{k=0}^\infty (-y)^k\right)^{1+2m-2n}(y^m-y^{m+1})
\end{align*}
does not contain $y^n$.

Therefore,
$$\frac{(1+y)^{2n+3}(1-y)}{((1+y)^2-2yx)^2}=\color{blue}{\underbrace{\text{something}}_{\text{“deg of $y$” $<n$}}}+\big((n+1) (2x)^n\big)y^n+\color{orange}{\underbrace{\text{something}}_{\text{“deg of $y$” $>n$}}},\qquad (|x|<\frac{(1+y)^2}{2|y|})$$
Take the $n$-th derivative with respect to $y$ and set $y\to 0$,
\begin{align*}
\left.\frac{\partial^n}{\partial y^n}\frac{(1+y)^{2n+3}(1-y)}{((1+y)^2-2yx)^2}\right|_{y=0}&=\color{blue}{0}+\big((n+1) (2x)^n\big)(y^n)^{(n)}+\color{orange}{0},\qquad (|x|<+\infty)\\
\left.\frac{\partial^n}{\partial y^n}\frac{(1+y)^{2n+3}(1-y)}{((1+y)^2-2yx)^2}\right|_{y=0}&=(n+1)! (2x)^n.\qquad (x\in \Bbb R)
\end{align*}