Note: The regions of convergence are somewhat different than you might think. Here I provide a complete Laurent expansion of $f$ around $z=0$. From this you should be able to check your results.
The function
\begin{align*}
f(z)&=\frac{1}{(z-1)(z-3)}\\
&=-\frac{1}{2}\frac{1}{z-1}+\frac{1}{2}\frac{1}{z-3}\\
\end{align*}
has two simple poles at $1$ and at $3$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $1$ and $3$ and see they determine three regions.
\begin{align*}
|z|<1,\qquad\quad
1<|z|<3,\qquad\quad
3<|z|
\end{align*}
The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the pole $1$ at the boundary of the disc. In the interior of this disc all two fractions with poles $1$ and $3$ admit a representation as power series at $z=0$.
The second region $1<|z|<3$ is the annulus with center $0$, inner radius $1$ and outer radius $3$. Here we have a representation of the fraction with pole $1$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $3 admits a representation as power series.
The third region $|z|>3$ containing all points outside the disc with center $0$ and radius $3$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.
A power series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}
=\frac{1}{a}\sum_{n=0}^\infty\left(-\frac{z}{a}\right)^n\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{a^{n+1}}z^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\left(-\frac{a}{z}\right)^n\\
&=\sum_{n=1}^{\infty}(-a)^{n-1}\frac{1}{z^n}
\end{align*}
We can now obtain the Laurent expansion of $f(x)$ at $z=0$ for all three regions
\begin{align*}
f(z)&=\frac{1}{2}\sum_{n=0}^{\infty}z^n
-\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{3^n}z^n\\
&=\frac{1}{2}\sum_{n=0}^{\infty}\left(1-\frac{1}{3^n}\right)z^n
\end{align*}
\begin{align*}
f(z)&=-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{z^n}
-\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{3^n}z^n
\end{align*}
\begin{align*}
f(z)&=-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{z^n}
+\frac{1}{2}\sum_{n=1}^{\infty}3^{n-1}\frac{1}{z^n}\\
&=\frac{1}{2}\sum_{n=1}^{\infty}\left(3^{n-1}-1\right)\frac{1}{z^n}\\
\end{align*}
Note: According to OPs comment let's assume a slightly different situation:
Challenge: Find the Laurent series expansion of $$f(z)=\frac{1}{(z-1)(z-3)}$$ at $z_0=3$ in the region $$0<|z-3|<2$$
We can immediately conclude from the representation of $f$ that $1$ and $3$ are simple poles.
Note, that in case of isolated singularities the radius of convergence is the distance from the center $z_0$ to the nearest singularity ($\neq z_0$).
Since $z_1=1$ is the nearest singularity to $z_0=3$ we have a radius of convergence $R=2$ and since $z_0$ is already a singularity we have a punctured disc as region of convergence
\begin{align*}
0<|z-3|<2
\end{align*}
When looking at
\begin{align*}
f(z)=-\frac{1}{2}\frac{1}{z-1}+\frac{1}{2}\frac{1}{z-3}
\end{align*}
we observe the term $\frac{1}{2}\frac{1}{z-3}$ is already the principal part of the Laurent expansion at $z_0=3$ and since $\frac{1}{2}\frac{1}{z-1}$ is analytic at $z_0=3$ we can expand it as power series.
\begin{align*}
f(z)&=-\frac{1}{2}\frac{1}{z-1}+\frac{1}{2}\frac{1}{z-3}\\
&=\frac{1}{2}\frac{1}{z-3}-\frac{1}{2}\frac{1}{(z-3)+2}\\
&=\frac{1}{2}\frac{1}{z-3}-\frac{1}{4}\frac{1}{1+\frac{z-3}{2}}\\
&=\frac{1}{2}\frac{1}{z-3}-\frac{1}{4}\sum_{n=0}^\infty \left(-\frac{1}{2}\right)^n(z-3)^n\\
&=\sum_{n=-1}^\infty \frac{(-1)^{n+1}}{2^{n+2}}(z-3)^n
\end{align*}
