Let $f(z)=\frac{\exp(z^2)}{z^3-1}$, $z\in\mathbb{C}$, and show that the Taylor series for $f$ around $0$ has the form $$ f(z)=\sum_{n=0}^\infty a_k z^k$$ where $a_0=-1$, $a_1=0$, $a_2=-1$, and $a_{2k+1}=a_{2k-2}$, $a_{2k+2}=a_{2k-1}-\frac{1}{(k+1)!}$ for $k\geq1$.
I found the Taylor series for $\exp(z^2)$ around $0$, and found that $$ f(z)=\frac{1}{z^3-1} \exp(z^2)=\frac{1}{z^3-1}\sum_{k=0}^\infty \frac{1}{k!} z^{2k}$$
but now I have no ideas on how to proceed. Any help appreciated.
Hint: $(z^3-1) f(z) = \exp(z^2)$. If $f(z) = \sum_k a_k z^k$, what are the coefficients of $z^{2k+1}$ and $z^{2k+2}$ in both sides of that equation?