Put $A=k[X,Y] \;$ . Consider the ideals $I=(X,Y)^n \subset A \; $ and $J=(f(X,Y) )\subset A$.
The ring you are interested in is $(A/J)/(I.A/J)$ . The trick is to replace it by the isomorphic ring $R:=(A/I)/(J.A/I)$.
[This is permitted because taking the quotient of a ring $A$ successively by two ideals $I,J$ does not depend on the order in which you perform the quotients: think about this!]
But then everything becomes clear: $A/I=k[X,Y]/(X,Y)^n \quad $ is local (see Dylan's comment) and so is any of its quotient, in particular our ring $R$.
(By the way: the irreducibility of $f$ is completely irrelevant.)