The following applies for some neighbourhood of zero, which is what we're interested in (the only possible singularity of the function within the unit circle) $\frac{1}{3\left(1+\frac{z}{3}\right)}\frac{1}{\sin 2z}e^z=\frac{1}{3}\left(1-\frac{z}{3}+\frac{z^2}{9}-...\right)\frac{1}{2z-\frac{8z^3}{3!}+...}\left(1+z+\frac{z^2}{2!}+...\right)=$ $=\frac{1}{3}\left(1-z+\frac{z^2}{9}-...\right)\,\frac{1}{2z}\frac{1}{1-\frac{4z^2}{3}+...}\left(1+z+\frac{z^2}{2!}+...\right)=$ $=\frac{1}{6z}\left(1-z+\frac{z^2}{9}+...\right)\left(1+z+\frac{z^2}{2}+...\right)\left(1+\frac{4z^2}{3}+...\right)=\frac{1}{6z}+...$ Thus, the function's residue at $\,z=0\,$ is $\,1/6\,$ and by the residue's theorem the integral's value is $\frac{1}{6}2\pi i=\frac{\pi i}{3} $
Of course, the residue of this pole (because that is what $\,z=0\,$ is: a simple pole, as can be easily checked) is waaaaaay easier to evaluate by the well-known formula $Res_{z=0}(f)=\lim_{z\to 0}\frac{ze^z}{(z+3)\sin 2z}=\frac{1}{6}$ but using power series can be serious fun...and pretty helpful sometimes if the pole's order is high.