If you allow singularitieslook at local rings, it is easy to construct such examples. For example, given any birational class if $X,Y$ are smooth of surfacessame dimension, you can find one with a rational double then for any point of fixed type (say $A_n, D_nx\in X,y\in Y$, E_6.E_7$or the completions of$E_8$). But these have only one analytic type, so taking different birational classesO_{X,x}, O_{Y,y}$ are isomorphic, you can find such but of course the algebraic local rings which are not necessarily isomorphic , but their completions are. It is somewhat more subtle to do the same if you fix the birational class and in general it is impossible except in the case of rational surfaces, when in some cases you can find such non-isomorphic ones. For (for example, if $R$ is the local ring at the origin of 3-space, $R/(x^5+y^3+z^2)$ and $R/(x^4y+x^5+y^3+z^2)$ both X,Y$are$E_8$singularities on a rational surface, non-isomorphicnot birational, but of course then even their completions fraction fields are isomorphic.not isomorphic.) 1 If you allow singularities, it is easy to construct such examples. For example, given any birational class of surfaces, you can find one with a rational double point of fixed type (say$A_n, D_n, E_6.E_7$or$E_8$). But these have only one analytic type, so taking different birational classes, you can find such local rings which are not isomorphic, but their completions are. It is somewhat more subtle to do the same if you fix the birational class and in general it is impossible except in the case of rational surfaces, when in some cases you can find such non-isomorphic ones. For example, if$R$is the local ring at the origin of 3-space,$R/(x^5+y^3+z^2)$and$R/(x^4y+x^5+y^3+z^2)$both are$E_8\$ singularities on a rational surface, non-isomorphic, but of course their completions are isomorphic.