@Arjun Sir

**0**^{n+m}1^{n+m}0^{n+m}^{ = }0^{n }0^{m }1^{m }1^{n }0^{n }0^{m }

Here for every 0^{n } we can push 0^{2n }(1 zero to macth 0 and 1 zero to match 1 )

for every 0^{m } we can push 0^{2m }, **So ultimately for every 0 we are pushing 2 0's to the stack **

For all 1^{m }pop 0^{m} (0's will be on top of the stack) ,for all 1^{n }pop 0^{n }(0's will be on top of the stack)

We are still left with **0**^{n+m }on our stack, which can be matched to last **0**^{n+m}^{ }

So isn't it CFL, What is wrong here Arjun Sir ?