Find $\int_{\vert z \vert = 1} {dz \over z(e^{1/ z}-1)}$ ${1 \over z(e^{1/ z}-1)}$ has a succession of isolated singularities $({1 \over 2k \pi} )_{k \in N}$ which converges to 0, then 0 is a non-isolated singularity. 
Edit: There is no residue on 0, what to do instead?
 A: Substituting $z\mapsto\frac1z$, $\frac{\mathrm{d}z}{z}\mapsto-\frac{\mathrm{d}z}{z}$ but the direction of the contour is reversed. Therefore,
$$
\begin{align}
\int_{|z|=1}\frac{\mathrm{d}z}{z\left(e^{1/z}-1\right)}
&=\int_{|z|=1}\frac{\mathrm{d}z}{z\left(e^z-1\right)}\\
&=-\pi i
\end{align}
$$
since
$$
\begin{align}
\frac1{z\left(e^z-1\right)}
&=\frac1{z^2\left(1+\frac z2+O\!\left(z^2\right)\right)}\\
&=\frac{1-\frac z2+O\!\left(z^2\right)}{z^2}\\
&=\frac1{z^2}\color{#C00000}{-\frac1{2z}}+O(1)
\end{align}
$$
A: 
METHODOLOGY $1$:

Exploiting Cauchy's Integral Theorem, the value of the integral is unaltered by integrating over $|z|=R>1$.  That is to say, 
$$\begin{align}
\oint_{|z|=1}\frac{1}{z(e^{1/z}-1)}\,dz&=\oint_{|z|=R>1}\frac{1}{z(e^{1/z}-1)}\,dz\\\\
&=\int_0^{2\pi}\frac{1}{Re^{i\phi}\left(e^{\frac{1}{Re^{i\phi}}}-1\right)}\,iRe^{i\phi}\,d\phi\\\\
&=i\int_0^{2\pi} \frac{1}{\frac{1}{Re^{i\phi}}\left(1+\frac{1}{2Re^{i\phi}}+O\left(\frac{1}{R^2e^{i2\phi}}\right)\right)}\,d\phi\\\\
&=i\int_0^{2\pi}Re^{i\phi}\left(1-\frac{1}{2Re^{i\phi}}+O\left(\frac{1}{R^2e^{i2\phi}}\right)\right)\,d\phi\\\\
&\to -i\pi\,\,\text{as}\,\,R\to \infty
\end{align}$$


METHODOLOGY $2$:

Alternatively, we can use the Residue at Infinity and write
$$\begin{align}
\oint_{|z|=1}\frac{1}{z(e^{1/z}-1)}\,dz&=-2\pi i \text{Res}\left(\frac{1}{z(e^{1/z}-1)},z=\infty\right)\\\\
&=-2\pi i \text{Res}\left(-\frac{1}{z^2}\frac{1}{z^{-1}(e^z-1)},z=0 \right)\\\\
&=2\pi i\left(-\frac12\right)\\\\
&=-i\pi
\end{align}$$


NOTES:

We can use Taylor's Theorem to write
$$\begin{align}
z(e^{1/z}-1)&=z\left(1+\frac{1}{z}+\frac{1}{2z^2}+O\left(\frac{1}{z^3}\right)\right)-z\\\\
&=\left(1+\frac{1}{2z}+O\left(\frac{1}{z^2}\right)\right)
\end{align}$$
Letting $z=Re^{i\phi}$, we obtain
$$\begin{align}
\frac{1}{z(e^{1/z}-1)}&=\frac{1}{\left(1+\frac{1}{2Re^{i\phi}}+O\left(\frac{1}{R^2e^{i2\phi}}\right)\right)}\\\\
&=\left(1+\frac{1}{2Re^{i\phi}}+O\left(\frac{1}{R^2e^{i2\phi}}\right)\right)^{-1}\\\\
&=\left(1-\frac{1}{2Re^{i\phi}}+O\left(\frac{1}{R^2e^{i2\phi}}\right)\right)
\end{align}$$
Then, with $dz=iRe^{i\phi}$, we arrive at the final lines.
