Solving $T(n)=4T(n/4) +(n/\log n)^2$
Let $H (m) := T(4^m)$. The recurrence is now $H(m) = 4 H(m-1) + 16^{m} / 4m^2$. 

By substituting, we get 

$4H(m-1) = 16 H(m-2) + 4\cdot16^{m-1} / 4(m-1)^2$

$16H(m-2) = 64 H(m-3) + 16\cdot16^{m-2} / 4(m-2)^2$

$64H(m-3) = 256 H(m-4) + 64\cdot16^{m-3} / 4(m-3)^2$

...

That sequence ends with $4^mH(0)$. So we need to add $16^{m} / 4m^2$, $4\cdot16^{m-1} / 4(m-1)^2$, $16\cdot16^{m-2} / 4(m-2)^2$, $64\cdot16^{m-3} / 4(m-3)^2$, ..., $4^mH(0)$. 

We extract a factor $16^{m} / 4$, then the numbers to add are $1 / m^2$, $1 / 4(m-1)^2$, $1 / 16(m-2)^2$, $1 / 64(m-2)^2$ etc. The sum is slightly larger than $16^m / 3m^2$. And T(n) is slightly larger than $(4/3)(n / log n)^2$.