To answer both your old and your new question at the very same time, we can consider a surprising relationship between power series and recurrence relations through generating functions. As a simple example, consider representing $\frac{1}{1-x}$ as a power series. In particular, we want to discover an $f_n$ such that
$$\frac{1}{1-x}=f_0+f_1x+f_2x^2+f_3x^3+\ldots$$
How do we do it? It proves pretty easy; let's multiply both sides by $(1-x)$ to obtain:
$$1=(1-x)(f_0+f_1x+f_2x^2+f_3x^3+\ldots)$$
Now, if we distribute the $(1-x)$ over the infinite sum, we get:
$$\begin{align}1=f_0 &+ f_1x + f_2x^2 + f_3x^3+f_4x^4+\ldots\\& -f_0x-f_1x^2-f_2x^3-f_3x^4-\ldots \end{align}$$
and doing the subtractions in each column, we get to the equation:
$$1=f_0+(f_1-f_0)x+(f_2-f_1)x^2+(f_3-f_2)x^3+\ldots$$
What's clear here? Well, every coefficient of $x$ has to be $0$ - so we get that $f_1-f_0$ and $f_2-f_1$ and $f_3-f_2$ must all be zero. In other words, $f_{n+1}=f_n$. Then, the constant term, $f_0$, must be $1$. Hence $f$ is defined as:
$$f_0=1$$
$$f_{n+1}=f_n.$$
That's a very simple recurrence relation, solved as $f_n=1$ meaning
$$\frac{1}{1-x-x^2}=1+x+x^2+x^3+\ldots$$
Okay, that's pretty cool, but let's try something harder, like setting
$$\frac{1}{1-x-x^2}=f_0+f_1x+f_2x^2+f_3x^3+\ldots$$
Now, if we multiply through by $(1-x-x^2)$ and expand like before, we get
$$\begin{align}1=f_0&+f_1x+f_2x^2+f_3x^3+\ldots\\&-f_0x-f_1x^2-f_2x^3-\ldots\\ &\qquad\,-f_0x^2-f_1x^3-\ldots\end{align}$$
and gathering terms again we get:
$$1=f_0+(f_1-f_0)x+(f_2-f_1-f_0)x^2+(f_3-f_2-f_1)x^3+\ldots+(f_{n+2}-f_{n+1}-f_n)x^{n+2}+\ldots$$
where we see again that $f_0=1$ and that $f_1-f_0$ is zero, hence $f_1=1$. Then, for higher terms we get that the coefficient of $x^{n+2}$ must be zero, hence
$$f_{n+2}-f_{n+1}-f_n=0$$
$$f_{n+2}=f_{n+1}+f_n$$
Oh hey, that's the Fibonacci sequence, so we get the identity
$$\frac{1}{1-x-x^2}=1+1x+2x^2+3x^3+5x^4+8x^5+13x^6+\ldots$$
Your equation will work the same way: Multiply through by the denominator, gather like terms, and then equate the coefficients on the left with those on the right. You will discover exactly the recurrence you found.
You don't even need differentiation or integration. $S(x)$ is well defined in the unit ball and:
$$ (1-x)\cdot S(x) = (1-x)\sum_{n=1}^{+\infty} n x^{n-1} = \sum_{n\geq 1}\left(n x^{n-1}-n x^{n}\right)=\sum_{n\geq 1}x^{n-1} = \frac{1}{1-x}$$
from which:
$$ S(x) = \frac{1}{(1-x)^2} $$
follows. As an alternative:
$$ S(x) = \frac{d}{dx}\sum_{n\geq 1}x^n = \frac{d}{dx}\left(\frac{1}{1-x}-1\right)=\frac{1}{(1-x)^2}.$$
Best Answer
$$\sum_{n=1}^\infty(-1)^{n+1}nx^{2n-2}=\sum_{n=1}^\infty n(-x^2)^{n-1}$$
$$=\frac{d}{dx}\sum_{n=1}^\infty(-x^2)^n$$
Now $\sum_{n=1}^\infty(-x^2)^n$ converges if $|-x^2|<1\iff -1<-x^2<1\iff1>x^2>-1$
If $x$ is real, $x^2\ge0>-1,$ we need $x^2<1\iff-1<x<1$