The definition of singularity in Wolfram MathWorld says "Complex singularities are points $z_0$ in the domain of a function $f$ where $f$ fails to be analytic." http://mathworld.wolfram.com/Singularity.html
But in the function $f(z)=\frac{1}{z}$, zero is a pole, thus a singularity. So according to the above definition, $z=0$ is in the domain of $f$?
No, $z=0$ is not in the domain of $f(z)=1/z$.
$z=0$ is a pole of $f$, and as in the case of every isolated singularity, $f$ is analytic in a neighbourhood of $z=0$ minus $\{0\}$.