The Laurent series expansion of $\cot z$ (from here) is
$$
\cot z = \dfrac{1}{z}-\dfrac{z}{3}-\dfrac{z^3}{45} + \Theta(z^5)
$$
For $z\in D(0, \pi)=\left\{z\in \mathbb{C} : \ |z-0| < \pi \right\}$
Then, the series of $z \cot z$ is given by
$$
f(z) = z \cdot \cot z = 1-\dfrac{z^2}{3}-\dfrac{z^4}{45} + \Theta(z^6)
$$
Then, it has a removable singularity at $z=0$.
Another way to do it is by computing the limit in $\mathbb{R}$
$$
\lim_{z \to 0} z \cdot \cot z = \lim_{z \to 0} \dfrac{z \cdot \cos z}{\sin z} = \underbrace{\lim_{z \to 0} \dfrac{z}{\sin z}}_{1} \cdot \underbrace{\lim_{z \to 0} \cos z}_{1} = 1
$$
As this limit exists, and $f(z)$ can be written as a Laurent series, then the limit in $\mathbb{C}$ is the same as the limit in $\mathbb{R}$.
EDIT: For the singularities at $z = k\pi$ for $k \in \mathbb{Z} \setminus \{0\}$, the singularity is of order one:
Using the result above we have that
$$
\lim_{z \to k\pi} f(z-k\pi) = \lim_{z \to 0} f(z) = 1
$$
And
\begin{align*}
\lim_{z \to k\pi} f(z) & = \lim_{z \to k\pi} z \cdot \dfrac{\cos z}{\sin z} \\ & = \lim_{z \to k\pi} \left[(z-k\pi) + k\pi\right] \dfrac{\cos z}{\sin z} \\ & = \underbrace{\lim_{z \to k\pi} \left(z-k\pi\right) \dfrac{\cos (z-k\pi)}{\sin (z-k\pi)}}_{1} + \underbrace{\lim_{z \to k\pi} k\pi \cdot \dfrac{\cos (z-k\pi)}{\sin (z-k\pi)}}_{\text{doesn't} \ \text{exist} \ \text{for} \ k\ne 0}
\end{align*}
Then, computing the limit of $(z-k\pi)f(z)$ for $k\ne 0$:
$$
\begin{align*}
\lim_{z \to k\pi} (z-k\pi)f(z) & = \lim_{z \to k\pi} (z-k\pi) z \cdot \dfrac{\cos z}{\sin z} \\ & = \lim_{z \to k\pi} \left[(z-k\pi)^2 + k\pi (z - k \pi)\right] \dfrac{\cos z}{\sin z} \\ & = \underbrace{\lim_{z \to k\pi} \left(z-k\pi\right)^2 \dfrac{\cos (z-k\pi)}{\sin (z-k\pi)}}_{0} + k \pi \underbrace{\lim_{z \to k\pi} (z-k\pi) \cdot \dfrac{\cos (z-k\pi)}{\sin (z-k\pi)}}_{1} \\
& = k\pi
\end{align*}
$$
Then $f(z)$ has one removable singularity at $z=0$ and one pole for $z=k\pi$ for $k \in \mathbb{Z} \setminus \{0\}$
