Proving an identity using formal power series 
4.
(a) Prove that $\dfrac{1-x^2}{1+x^3}=\dfrac{1}{1+\frac{x^2}{1-x}}$.
(b) By expanding each side of the identity in (a) as a power series, and considering the coefficient of $x^N$, prove that
  $$\left|\sum_{k\geq0}(-1)^{k}{N-k-1 \choose N-2k}\right|=\begin{cases}
0 & \text{ if }N\equiv1\pmod3\\
1 & \text{ otherwise }
\end{cases}$$

Apparently I need to use formal power series. However, I can't seem to derive anything useful by attempting to divide $\frac{1-x^2}{1+x^3}$. How should I proceed? More generally, how should I go about proving identities using formal power series, and how should I expand a fraction into a power series?

For the first, $$ \frac{1}{1 + \frac{x^2}{1 - x}} = \frac{1 - x}{1 - x + x^2} = \frac{(1 + x)(1 - x)}{(1 + x)\left(1 - x + x^2\right)} = \frac{1 - x^2}{1 + x^3} $$
For the second, consider the Maclaurin Series of $ \frac{1 - x^2}{1 + x^3} $ which is $$ \left( 1 - x^2\right)\sum_{k = 0}^\infty \left(-x^3\right)^n = \sum_{k = 0}^\infty a_nx^n$$ where $$ a_n = \begin{cases} &0 & x \equiv 1 \mod 3 \\ -&1 & x \equiv 2 \mod 6 \vee x \equiv 3 \mod 6 \\ &1 &x \equiv 5 \mod 6 \vee x \equiv 0 \mod 6 \end{cases} $$
Meanwhile, he Maclaurin Series for $ \frac{1}{1 + \frac{x^2}{1 - x}} $ is $ \displaystyle \sum_{n = 0}^\infty f(n) x^n $ where $$ f(n) := \sum_{k\ge 0 } (-1)^k  \dbinom{n -k -1}{n - 2k} $$Because the two functions are equivalent near $ x = 0 $, their Maclaurin series are the same and the equality follows.
The hard part is proving that definition of $ f(n) $. It involves expanding $ \frac{x^2}{1 - x} $ as its own Maclaurin series and then using a geometric series on $ \frac{1}{1 - \left(-\frac{x^2}{1 - x}\right)}.
