This function will be continuous at 0 and 1 and no other points. To see this intuitively, you can just note that x² and x³ are unequal for x≠0,1, so your function will be somehow "jumping around" near these points.

To prove continuity at 0, you can simply note that as x-->0, both x² and x³ go to zero. Therefore f(x) also goes to 0 as x-->0. Also f(0)=0 so we have continuity at 0 (if you want an epsilon-delta proof, just pick delta<min{sqrt(epsilon),1} so |delta²| and |delta³| are both <epsilon). Continuity at 1 is basically the same.

To prove discontinuity away from 0,1, let a≠0,1 be our fixed point of consideration. Probably the easiest way forward is to find sequences (q_n) and (r_n) of rational and irrational numbers respectively, which both converge to a (existence of such sequences follows as Q and R\Q are dense in R). Then, if our function f were continuous at a, we'd have f(q_n) and f(r_n) both converging to f(a). But these limits are a³ and a² respectively, which are unequal. Contradiction so f is not continuous at a