Given the function $$f(z) = z^2 \cdot \frac{{e^{1/z}}}{{z-1}}$$, we want to find its Laurent series expansion around the point $z = 0$. This is my approach. I am not sure about the ending. First, let's consider the function $$g(z) = \frac{{e^{1/z}}}{{z-1}}.$$ We can expand $g(z)$ in a Taylor series as follows:
$$ g(z) = \sum_{n=0}^{\infty} \left(\sum_{k=0}^{\infty} \frac{1}{{(n+k)!}}\right) \left(\frac{1}{z}\right)^n $$
Now, let's multiply $g(z)$ by $z^2$:
$$ z^2 \cdot g(z) = \sum_{n=0}^{\infty} \left(\sum_{k=0}^{\infty} \frac{1}{{(n+k)!}}\right) z^{2-n} $$
To obtain the Laurent series, we need negative powers of $z$. Let's rewrite the series in a more convenient form:
$$ z^2 \cdot g(z) = \sum_{n=-\infty}^{\infty} \left(\sum_{k=0}^{\infty} \frac{1}{{(n+1+k)!}}\right) z^n $$
Therefore, the Laurent series expansion of $$f(z) = z^2 \cdot \frac{{e^{1/z}}}{{z-1}}$$ around $z = 0$ is:
$$ f(z) = \sum_{n=-\infty}^{\infty} \left(\sum_{k=0}^{\infty} \frac{1}{{(n+1+k)!}}\right) z^n $$