Is this language context free? In a recent test, I was asked to recognize if the below language is context  free:

$\qquad\displaystyle L = \{0^{n+m}1^{n+m}0^m \mid n,m \geq 0\}$ 

I think it is context free, and can be accepted by below context free grammar, where $S$ is the start symbol and $Y$ is a non-terminal:

$\qquad S \to S0 \mid Y$

$\qquad Y \to 0Y1 \mid \epsilon$

However, my answer was considered wrong and that the language $L$  is not context free.

I'm confident about my answer, but the response has got me confused. Is my understanding correct? Please let me know if I've missed something.
You can use Ogden's lemma. Choose the word $w = 0^p1^p0^p$ for large enough $p$, and mark the rightmost $0^p$. Ogden's lemma gives you a decomposition $w = uxyzv$ with $xz$ pumpable and containing at least one marked point. Since $xz$ contains a $0$, it can't contain a $1$, as otherwise $ux^2yz^2v \notin 0^*1^*0^*$. A simple case analysis now leads to a contradiction.