What you have done up to now is correct. The idea is that you should prove that the 'pumping' fails for all possible partitions of $w$. You have considered just two families of partitions, namely, the one for which $vxy\in0^+$ and the one for which $vxy\in1^+$.

You are missing the cases in which the sliding window (ie. $vxy$) crosses the boundary between $0$ and $1$.

Hence, next step is to consider the following cases (here $|w|$ is the length of $w$):

1. $vxy=0^{|vx|}0^h1^k$ where $h,k\geq 1$ and $h+k+|vx|\leq p$

2. $vxy=0^{|vx|}1^{|y|}$ where $|y|>0$ and $|vxy|\leq p$

3. $vxy=0^{|v|}0^h1^k1^{|y|}$ where $h+k=|x|>0$ and $|vxy|\leq p$

4. $vxy=0^{|v|}1^{|xy|}$ where $|v|>0$, $|xy|>0$ and $|vxy|\leq p$

All of the can be easily solved with the technique you used with the former cases.

What you have done up to now is correct. The idea is that you should prove that the 'pumping' fails for all possible partitions of $w$. You have considered just two families of partitions, namely, the one for which $vxy\in0^+$ and the one for which $vxy\in1^+$.

You are missing the cases in which the sliding window (ie. $vxy$) crosses the boundary between $0$ and $1$.

Hence, next step is to consider the following cases (here $|w|$ is the length of $w$):

1. $vxy=0^{|vx|}0^h1^k$ where $h,k\geq 1$ and $h+k+|vx|\leq p$

2. $vxy=0^{|vx|}1^{|y|}$ where $|y|>0$ and $|vxy|\leq p$

3. $vxy=0^{|v|}0^h1^k1^{|y|}$ where $h+k=|x|>0$ and $|vxy|\leq p$

4. $vxy=0^{|v|}1^{|xy|}$ where $|v|>0$, $|xy|>0$ and $|vxy|\leq p$

All of the above cases can be easily solved with the technique you used with the former ones.