Answer
$-x-\frac{3x^2}{2}-\dfrac{4x^3}{3}-x^4$
Work Step by Step
We have the Maclaurin series for $ \sin x$ and $e^x$ as follows
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+$$
$$\ln (1-x)=-x+\frac{x^2}{2!}-\frac{x^3}{3!}-\frac{x^4}{4!}+$$
Now, we have the Maclaurin series of $e^x \ln(1-x)$ as:
$$e^x\ln (1-x)=\\
=(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+)(-x+\frac{x^2}{2!}-\frac{x^3}{3!}-\frac{x^4}{4!}+..)\\
=-x-\frac{3x^2}{2}-\dfrac{4x^3}{3}-x^4$$
