$\int_{|z| = \frac{1}{2}} \frac{e^{1/z}}{1-z}$
Singularities:
$1-z = 1$ implies $z=1$ is a simple pole. It is not in $|z| = \frac{1}{2}$.
Whereas, $e^{1/z}$ has an essential singularity inside $|z| = \frac{1}{2}$.
Consider $\frac{e^{1/z}}{1-z}= (1+z+z^2+...)(1+\frac{1}{z}+\frac{1}{2!z^2} + ...)$
It seems that the coefficient of $1/z = e-1$.
Therefore, $\int_{|z| = \frac{1}{2}} \frac{e^{1/z}}{1-z} = 2i\pi(e-1)$.
Am I right?
We could show ( observe that we take the Cauchy product, i.e. according to powers of $\;z\;$)
$$\frac{e^{1/z}}{1-z}=\sum_{n=0}^\infty z^n\sum_{k=0}^\infty\frac1{k!z^k}=\sum_{m=0}^\infty\frac{z^k}{(m-k)!z^{m-k}}=\sum_{m=0}^\infty\frac{z^{2k-m}}{(m-k)!}$$
we now observe that
$$2k-m=-1\iff m=2k+1\implies\,\text{in the above development, the wanted coefficient is}$$
$$\sum_{m=0}^\infty\frac1{(k+1)!}=\sum_{n=1}^\infty\frac1{n!}=e-1$$