I want to find out, what kind of singularties does
$ f(z)= \frac{1}{z^3-z^5}$ have.
I would do the following steps: $ f(z)= \frac{1}{z^3-z^5} = \frac{1}{z^3(1-z)(1+z)}$ so I have $ z_1=1, z_2=-1 , z_3=0$ I can argue by the limit test, that $ z_i$ are poles. For example $ \lim_{z\rightarrow 1} (1-z)f(z)= \lim_{z\rightarrow 1} \frac{1}{z^3(1+z)}= \frac{1}{2} \ne 0 $
Am I right?
How can I find the Laurent series of f and classify the singularties without using the Limit test.
You are right: they are all poles.
Now, consider the point $0$. Since$$f(z)=\frac1{z^3}\times\frac1{1-z^2}$$and since, near $0$, you have$$\frac1{1-z^2}=1+z^2+z^4+z^6+\cdots$$then (again, near $0$) you have$$f(z)=\frac1{z^3}+\frac1z+z+z^3+\cdots$$and this confirms that $0$ is a pole.
You can do a similar thing at $\pm1$.