Approximating a series for asymptotic upper bound
To make things a bit simpler, let us assume that we use base 4 logarithm. Following your steps, for $n = 4^k$ and $T(1) = 0$ we get
$$
\begin{align*}
T(n) &amp;= \frac{n^2}{(\log_4 n)^2} + 4\frac{(n/4)^2}{(\log_4 n-1)^2} + 16\frac{(n/16)^2}{(\log_4 n - 2)^2} + \cdots \\ &amp;=
\frac{n^2}{(\log_4 n)^2} \left[ 1 + \frac{1}{4(1-1/\log_4 n)^2} + \frac{1}{16(1-2/\log_4 n)^2} + \cdots \right].
\end{align*}
$$
Let us now use the series expansion
$$
\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + \cdots.
$$
Substituting this in the expression above, we get
$$
\begin{align*}
\frac{T(n)}{n^2/\log^2_4n} &amp;= 1 + \frac{1}{4} + \frac{1}{16} + \cdots \\ &amp;+
2\left[ \frac{1}{4\log_4 n} + \frac{2}{16\log_4 n} + \cdots \right] \\ &amp;+
3 \left[ \frac{1}{4\log_4^2n} + \frac{2^2}{16\log_4^2n} + \cdots \right] + \cdots \\ &amp;\approx \frac{4}{3} + \frac{8/9}{\log_4 n} + \frac{20/9}{\log_4^2 n} + \cdots
\end{align*}
$$
The error in the approximation is polynomially small: for example 
$$
\sum_{k=0}^{\log_4 n-1} \frac{1}{4^k} = \frac{4}{3} - \frac{4/3}{4^{\log_4 n}} = \frac{4}{3} - \frac{4/3}{n}.
$$
The error is somewhat larger for the other summands, but still polynomially small. Taylor's theorem (or direct calculation) shows that if we truncate the series for $1/(1-x)^2$ after the term $(d+1)x^d$ then we get an error of $O(x^{d+1})$. Altogether, this shows that the following asymptotic statements hold:
$$
\begin{align*}
T(n) &amp;= \frac{4}{3} \frac{n^2}{\log_4^2 n} + O\left(\frac{n^2}{\log_4^3 n}\right), \\
T(n) &amp;= \frac{4}{3} \frac{n^2}{\log_4^2 n} + \frac{8}{9} \frac{n^2}{\log_4^3 n} + O\left(\frac{n^2}{\log_4^4 n}\right), \\
T(n) &amp;= \frac{4}{3} \frac{n^2}{\log_4^2 n} + \frac{8}{9} \frac{n^2}{\log_4^3 n} + \frac{20}{9} \frac{n^2}{\log_4^4 n} + O\left(\frac{n^2}{\log_4^5 n}\right), \ldots \\
\end{align*}
$$