I think some points of confusion can be clarified by looking at power series first.
Let $B(z_0,r) = \{z | |z-z_0|<r\}$, $\operatorname{ann}(z_0,r_1,r_2)= \{ z | r_1 < |z-z_0| < r_2\}$.
A power series centred at $z_0$ given by $z \mapsto \sum_{n=0}^\infty a_n (z-z_0)^n$ defines an analytic function for $z \in B(z_0,R)$, and diverges for $|z-z_0|>R$, where $\frac{1}{R} = \limsup_n \sqrt[n]{|a_n|}$. (Hence $R$ is the largest number such that the power series is analytic on $z \in B(z_0,R)$).
If the function defined by the power series has a non-removable singularity at some $z_1$, then clearly $R\le |z_1-z_0|$ (otherwise the function would be analytic at $z_1$). This is also what limits the radius of convergence of the Laurent series.
The power series coefficients (and hence the radius of convergence) depend on the function and the choice of $z_0$. For example, take $f(z) = \frac{1}{z}$. If we take $z_0=1$, then $f(z) = 1 - (z-1)+(z-1)^2-\cdots$ on $z \in B(1,1)$, whereas if we take $z_0=2$, we have $f(z) = 1-\frac{z-2}{2}+(\frac{z-2}{2})^2-\cdots$ on $z \in B(2,2)$. So $f$ has different power series in different regions. In this example, the radius of convergence of the power series will always be the distance from the centre to the singularity at $z=0$.
If a function $f$ is analytic on $\operatorname{ann}(z_0,r_1,r_2)$, then it has a Laurent series that converges uniformly & absolutely on any compact subset of the annulus. As with the power series, we must have ${r_2} \le {1 \over \limsup_{n \to \infty} \sqrt[n]{|a_n|}}$ and ${r_1} \ge { \limsup_{n \to \infty} \sqrt[n]{|a_{-n}|}}$.
By separating the negative and non-negative coefficient index parts of the Laurent expansion, we see that $f$ can be written as $f(z) = g_1(\frac{1}{z-z_0})+g_2(z-z_0)$, where $g_1,g_2$ are functions that are analytic on $B(0,\frac{1}{r_1})$ and $B(0,r_2)$ respectively. Hence the above remarks regarding power series can by applied mutatis mutandis to the Laurent series, taking into account the two-sided nature of things.
Note that all that is required is that $f$ be analytic on $\operatorname{ann}(z_0,r_1,r_2)$. There is no requirement that the center of the annulus lie on a singularity. For example, with $f(z) = \frac{1}{z}$ above, $f$ is analytic on
$\operatorname{ann}(2,1,2)$, and the Laurent expansion is precisely the power series expansion in this case.
As with the power series, the Laurent expansion depends on the chosen annulus.
As @DanielFischer told you, the regions to consider are $(1)$ $\lvert z\rvert<1$, $(2)$ $1<\lvert z\rvert < 2$, and $(3)$ $\lvert z\rvert >2$. For $(1)$, your solution is correct.
For $(2)$, we want Geometric series with $\lvert r\rvert <1$ so $1<\lvert z\rvert\Rightarrow \frac{1}{\lvert z\rvert} < 1$ and $\frac{\lvert z\rvert}{2}<1$. Then
\begin{align}
\frac{1/3}{z+1}+\frac{2/3}{z-2} &=\frac{1}{z}\frac{1/3}{1+\frac{1}{z}}-\frac{1/3}{1-\frac{z}{2}}\\
&=\frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{1}{z}\Bigr)^n-\frac{1}{3}\sum_{n=0}^{\infty}\Bigl(\frac{z}{2}\Bigr)^n\\
&=\frac{1}{3}\sum_{n=0}^{\infty}\biggl[(-1)^n\Bigl(\frac{1}{z}\Bigr)^{n+1}-\Bigl(\frac{z}{2}\Bigr)^n\biggr]
\end{align}
For $(3)$, we have $\lvert z\rvert >2$ so $\frac{1}{\lvert z\rvert} < 1$ which we found in $(2)$ and $\frac{2}{\lvert z\rvert}<1$. We only need to determine the Laurent series for $\frac{2}{\lvert z\rvert}<1$.
\begin{align}
\frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{1}{z}\Bigr)^n+\frac{2/3}{z-2}&=\frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{1}{z}\Bigr)^n+\frac{1}{z}\frac{2/3}{1-\frac{2}{z}}\\
&=\frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{1}{z}\Bigr)^n+\frac{2}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{2}{z}\Bigr)^n\\
&=\frac{1}{3}\sum_{n=0}^{\infty}(-1)^n\biggl[\Bigl(\frac{1}{z}\Bigr)^{n+1}+\Bigl(\frac{2}{z}\Bigr)^{n+1}\biggr]
\end{align}
Best Answer
Yes, the singulaty at $z = 0$ is due ot $z^2$
Try to draw the regions $0 < |z| < 3$ and $|z| > 3$. And also notice that $\frac{3}{|z|} < 1$ so we may use Taylor series around the point zero.
The series are as follow:
For $0 < |z| <3 $ we have that
$$f(z)=\frac{1}{z^2(z-3)} = \frac{1}{z^2}\frac{1}{(z-3)} = -\frac{1}{3z^2}\frac{1}{(1-\frac{z}{3})} = -\frac{1}{3z^2} \sum_{n=0}^{\infty} \frac{z^n}{3^n} =- \sum_{n=0}^{\infty} \frac{z^{n-2}}{3^{n+1}}$$
For $|z| > 3$ we have that
$$f(z)=\frac{1}{z^2(z-3)} = \frac{1}{z^2}\frac{1}{(z-3)} = \frac{1}{z^3}\frac{1}{(1-\frac{3}{z})} = \frac{1}{z^3} \sum_{n=0}^{\infty} \frac{3^n}{z^n} = \sum_{n=0}^{\infty} \frac{3^n}{z^{n+3}}$$
Any questions leave in the comments.