You seek a solution $\rho$ of the equation $f'(\rho)=0$, hence
$$\rho^2=1+w^{-2}\rho^{2p}.$$
The solution should remain $>0$ when $w\rightarrow\infty$.    
<sub>The OP says the solution should vanish as $1/w$, but that is mistaken, I think.</sub>   

To gain some insight, take $p=2$, then the solution is 
$$\rho=\frac{\sqrt{w^2-w \sqrt{w^2-4}}}{\sqrt{2}},$$
which goes to $1$ when $w\rightarrow\infty$. The critical point $w_c$ is the smallest $w$ for which this solution exists, which is $w_c=2$.

For general $p$, the calculation of $w_c$ proceeds as follows. Define $u=1/w^2$ and $T=\rho^2$, then 
$$u=T^{1-p}-T^{-p}.$$
The critical $u_c=w_c^{-2}$ is reached when $dT/du\rightarrow\infty$ (the location of the square root singularity), hence $du/dT=0$ which gives $T=p/(p-1)\Rightarrow u_c=(p-1)^{p-1}/p^p$, and thus
$$w_c=p^{p/2} (p-1)^{(1-p)/2}.$$