A function #f# is even if #f(-x) = f(x)# and odd if #f(-x) = -f(x)#.
In this case, though, for any #x in RR# where #x!=0#, we have that one of #sqrt(x)# and #sqrt(-x)# is real, and the other is imaginary. Thus, in all such cases, #3sqrt(-x)!=3sqrt(x)# and #3sqrt(-x)!=-3sqrt(x)#, meaning #3sqrt(x)# is neither even nor odd.