(Cross-post from math.stackexchange, where it has received no attention.)

Orbifolds $\mathbb{C}^2/\mathbb{Z}_n$, given by the action $(x, y) \mapsto (\zeta x, \zeta^{-1} y)$ with $\zeta$ a primitive $n^\text{th}$ root of unity, admit smooth deformations. This is because they are hypersurface singularities; there are three independent invariants $u_1 = x^n$, $u_2 = y^n$, and $u_3 = xy$, and they satisfy one relation: $u_1 u_2 - u_3^n = 0$. Deforming to $u_1 u_2 - u_3^n + t = 0$ gives a flat family, and the fibres away from $t=0$ are smooth.

What about other orbifolds, like $(x, y) \mapsto (\zeta x, \zeta y)$? Is there an easy way to check whether they are smoothable?

(Some context: I know that not all orbifolds are smoothable, but don't really understand why. For example, it seems to be common knowledge that the orbifold $\mathbb{C}^3/\mathbb{Z}_3$, given by the action $(x, y, z) \to (\zeta x, \zeta y, \zeta z)$ where $\zeta = e^{2\pi i/3}$, is rigid. The invariant ring has 10 generators, satisfying 27 relations, and I have no idea how to go about showing that this admits no deformations.)