If $p$ is a proposed pumping length, let $q > p$ be prime, and consider the string $w = a^qb^{(q^2)} \in L$. (Clearly $|w| = q+q^2 \geq p$.)
Let $w = xyz$ be a decomposition such that
- $y \neq \varepsilon$, and
- $|xy| \leq p$.
In follows that $y = a^k$ for some $1 \leq k \leq p < q$, and so "pumping" it zero times results in the string $$ xy^0z = xz = a^{q-k}b^{(q^2)}$$ where $1 < q-k < q$. As the only factors of $q^2$ are $1,q,q^2$ (remember that $q$ is prime) it follows that $\operatorname{gcd} (q-k,q^2) = 1$, so $xy^0z$ is not in $L$.
Your basic idea is fine, but some of the details are wrong.
Yes, you can use the string $s=0^p1^{2p}0^p$. You could also use the string $0^p1^{p+1}0^p$: $L$ can also be described as the set of all words $0^k1^\ell0^k$ such that $0<k<\ell$. And when you split $s$ as $s=xyz$ with $|xy|\le p$ and $|y|>0$, it is true that $xy$ must be contained in the first block of zeroes. However, you cannot conclude that $x=\epsilon$ and $y=0^p$. All that you can conclude is that there are non-negative integers $r$ and $s$ such that $x=0^r$, $y=0^s$, $r+s\le p$, and $s\ge 1$. Then $z=0^{p-r-s}1^{2p}0^p$, and
$$xy^kz=0^r0^{ks}0^{p-r-s}1^{2p}0^p=0^{p+(k-1)s}1^{2p}0^p$$
for all $k\ge 0$. This word is in $L$ if and only if the two blocks of zeroes are the same size, i.e, if and only if $p+(k-1)s=p$, which is the case precisely when $k=1$, so pumping the word up (by making $k>1$) or down (by making $k=0$) will give you a word not in $L$. In particular, you can — as you did — take $k=2$ to get
$$xy^2z=0^{p+s}1^{2p}0^p\notin L\,,$$
since $p+s\ne p$, and conclude that $L$ is not regular.
Best Answer
Let $p$ be a prime number greater than the pumping length, and try $0^{p-1}1^p$. Show that when you pump up, you get something of the form $0^{p-1+kr}1^p$ for some $r$ such that $1\le r<p$, and show that there is a $k\ge 1$ such that $kr-1$ is a multiple of $p$, i.e., such that $kr\equiv 1\pmod p$.