Pumping Lemma: Proving L={0^n 1^m |m ≤n^2} is not a CFL
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 $|y|&gt;0$ and $|vxy|\leq p$

3) $vxy=0^{|v|}0^h1^k1^{|y|}$ where $h+k=|x|&gt;0$ and $|vxy|\leq p$

4) $vxy=0^{|v|}1^{|xy|}$ where $|v|&gt;0$, $|xy|&gt;0$ and $|vxy|\leq p$

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