Take a string $w$ which belongs to that language $L$.
Let, $w$ be $01^k==01^k$.
Dividing $w$ into $xyz$ where $|xy|$ is our pumping length and $|y|>=1.$
Let, $x=0, y=1^i, z=(1^{k-1}==1^k0)$,
Then, $w=(x.y.z)=(0.1.1^{k-1}==1^k0)$
According to the pumping lemma if $y$ is pumped $i$ times, then for all $i$ it must also belong to $L$.
So, $xy^iz=(0.(1)^i.(1)^{k-1}==1^k0)$.
You can see for except for $i=1$, $xy^iz$ will not belong to $L$. So, $L$ must not be Regular.