Let $U\subset\mathbb{C}$ be open and $f:U\setminus\{z_0\}\rightarrow\mathbb{C}$ holomorphic with a pole of order $k$ in $z_0$. Further let $G\subset\mathbb{C}$ be a domain and $\varphi:G\rightarrow\mathbb{U}$ holomorphic and not constant with $\varphi(w_0)=z_0$.
What can you say about the type of the singularity at $w_0$ of $f\circ \varphi$?
The requorements on $\varphi$ imply that $\varphi$ is a open map.
First, note the condition that $\phi$ is nonconstant is necessary so that $f\circ \phi$ is defined in some neighborhood of $w_0$ (since by analyticity there is a neighborhood in which $\phi(w)=z_0$ only for $w=w_0$), so that we're speaking of an isolated singularity.
The composition $f \circ \phi$ has a pole at $w_0$ if and only if $\lim_{w \to w_0} |(f \circ \phi)(w)| =\infty$. By continuity $\lim_{w \to w_0} \phi(w) = z_0$, and since $f$ has a pole at $z_0$ $\lim_{w \to w_0} |(f \circ \phi)(w)| = \lim_{z \to z_0} |f(z)| = \infty$, and thus $\lim_{w \to w_0} |(f \circ \phi)(w)| $ is indeed $\infty$ and $f \circ \phi$ has a pole at $w_0$.
Note we can't describe the order of the pole without knowing more about $\phi$: for example we can take $f(z)= 1/z^k$ and $\phi(z)=z^n$ to get a pole of order $kn$.