The language is L = {$1^{n}01^{n} \dotplus 1$, n>=1}
Notice here we need to check number of 1's before 0 is equal to number of 1's after zero. Which we can not achieve by DFA. So it is not regular.
This language is CFL as well as DCFL, At start we push all the 1's onto the stack, when we see 0 we ignore and after every 1 we pop 1's from the stack, if we end up with empty stack, we accept this language. So it is CFL as well as DCFL