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I want to find the first terms of the Laurent's series in $z=-1$

$$f(z)=\frac{e^{\frac{1}{z}}}{z^{3}+1}$$ $$f(z)=\frac{e^{\frac{1}{z}}}{(z+1)(z^{2}-z+1)}$$ $$e^{\frac{1}{z}}=\frac{1}{e}-\frac{z+1}{e}-\frac{(z+1)^{2}}{2e}-\frac{(z+1)^{3}}{6e}+O(z+1)^{4}$$ That's ok. I have some problems with the following espression: $$f(z)=\frac 1 {(z+1)(z^{2}-z+1)}=\frac{1}{z+1}\, \, \, \, \frac{1}{z^{2}-z+1}= $$$$\frac{1}{z+1} \left ( \frac{1}{3}+\frac{z+1}{3}+\frac{2}{9}(z+1)^{2}+\frac{1}{9}(z+1)^{3}+O(z+1)^{4} \right )$$ From Mathematica I know that the first terms of the series are the ones I have written above but I don't know how to get them.Someone can show me and explain the steps? Thank you so much!

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    Try looking up Cauchy products. Personally, if I only needed the first term, I'd multiply everything by $(z+1)$, then take the limit as $z\to-1$, and then divide by $(z+1)$ at the end. – 2017-02-27

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The given problem is equivalent to finding the Laurent expansion at the origin for $\frac{e^{\frac{1}{t-1}}}{(t-1)^3+1}$.
We have: $$\exp\left(\frac{1}{t-1}\right) = \frac{1}{e}\exp\left(-t-t^2-t^3-\ldots\right)=\frac{1}{e}-\frac{t}{e}-\frac{t^2}{2e}-\frac{t^3}{6e}+o(t^3)\tag{1}$$ and $$ \frac{1}{(t-1)^3+1} = \frac{1}{t^3-3t^2+3t} = \frac{1}{t}\cdot\frac{1}{3-3t+t^2} = \frac{1}{3 t}+\frac{1}{3}+\frac{2 t}{9}+\frac{t^2}{9}+\frac{t^3}{27}+o(t^3)\tag{2}$$ hence, by Cauchy's product,

$$ \frac{1}{(t-1)^3+1}\,\exp\left(\frac{1}{t-1}\right) = \frac{1}{3 e t}-\frac{5 t}{18 e}-\frac{t^2}{3 e}+o(t^3).\tag{3}$$