$z\cdot e^{1/z}\cdot e^{-1/z^2}$ at $z=0$.
My answer is removable singularity. $ \lim_{z\to0}\left|z\cdot e^{1/z}\cdot e^{-1/z^2}\right|=\lim_{z\to0}\left|z\cdot e^{\frac{z-1}{z^2}}\right|=\lim_{z\to0}\left|z\cdot e^{\frac{-1}{z^2}}\right|=0. $ But someone says it is an essential singularity. I don't know why.