= Solution
The <Maclaurin series> give $e^{-x^3}=1-x^3+O(x^6)$ and $\cosh x=1+x^2/2+x^4/24+O(x^6)$. Inverting the latter using the <geometric series> gives
$$
\frac1{\cosh x}=1-\frac{x^2}{2}+\left(\frac14-\frac1{24}\right)x^4+O(x^6)
=1-\frac{x^2}{2}+\frac5{24}x^4+O(x^6).
$$
Multiplying the two expansions yields
$$
\boxed{\frac{e^{-x^3}}{\cosh x}=1-\frac{x^2}{2}-x^3+\frac5{24}x^4+O(x^5).}
$$
The remainder is $O(x^5)$ because the product already generates an $x^5$ term from $(-x^3)(-x^2/2)$.
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