Solution
= Solution
Use the <binomial series> and exponential <Maclaurin series>:
$$
(1+x/3)^{-3}=1-x+\frac23x^2-\frac{10}{27}x^3+O(x^4),\qquad
1-e^x=-x-\frac{x^2}2-\frac{x^3}6-\frac{x^4}{24}+O(x^5).
$$
The <Cauchy product> is
$$
(1-e^x)(1+x/3)^{-3}=-x+\frac{x^2}2-\frac{x^3}3+\frac{35x^4}{216}+O(x^5).
$$
Adding $\log(1+x)=x-x^2/2+x^3/3-x^4/4+O(x^5)$ cancels the first three powers. Hence the first nonzero term is
$$
\boxed{-\frac{19}{216}x^4.}
$$