Since this question is asked frequently I will try to work out a
solution for generic positive integers $a$ where $a\ge 2$.
Suppose we have $T(0)=0$ and
$$T(1)=T(2)=\ldots =T(a-1)=1$$
and for $n\ge a$
$$T(n) = a T(\lfloor n/a \rfloor) + n \lfloor \log_a n \rfloor.$$
Furthermore let the base $a$ representation of $n$ be
$$n = \sum_{k=0}^{\lfloor \log_a n \rfloor} d_k a^k.$$
Then we can unroll the recurrence to obtain the following exact
formula for $n\ge a$
$$T(n) = a^{\lfloor \log_a n \rfloor}
+ \sum_{j=0}^{\lfloor \log_a n \rfloor -1}
a^j \times (\lfloor \log_a n \rfloor - j) \times
\sum_{k=j}^{\lfloor \log_a n \rfloor} d_k a^{k-j}.$$
Now to get an upper bound consider a string consisting of the digit
$a-1$ to obtain
$$T(n) \le a^{\lfloor \log_a n \rfloor}
+ \sum_{j=0}^{\lfloor \log_a n \rfloor -1}
a^j \times (\lfloor \log_a n \rfloor - j) \times
\sum_{k=j}^{\lfloor \log_a n \rfloor} (a-1) \times a^{k-j}.$$
This simplifies to
$$a^{\lfloor \log_a n \rfloor}
+ \sum_{j=0}^{\lfloor \log_a n \rfloor -1}
a^j \times (\lfloor \log_a n \rfloor - j) \times (a-1)
\sum_{k=0}^{\lfloor \log_a n \rfloor-j} a^k$$
which is
$$a^{\lfloor \log_a n \rfloor}
+ \sum_{j=0}^{\lfloor \log_a n \rfloor -1}
a^j \times (\lfloor \log_a n \rfloor - j)
(a^{\lfloor \log_a n \rfloor + 1 -j} -1)$$
which turns into
$$a^{\lfloor \log_a n \rfloor}
+ \sum_{j=0}^{\lfloor \log_a n \rfloor -1}
(\lfloor \log_a n \rfloor - j)
(a^{\lfloor \log_a n \rfloor + 1} - a^j).$$
The sum produces four terms.
The first is
$$\lfloor \log_a n \rfloor^2
a^{\lfloor \log_a n \rfloor + 1}.$$
The second is
$$- \lfloor \log_a n \rfloor
\frac{a^{\lfloor \log_a n \rfloor}-1}{a-1}.$$
The third is
$$- \frac{1}{2} a^{\lfloor \log_a n \rfloor + 1}
(\lfloor \log_a n \rfloor -1) \lfloor \log_a n \rfloor$$
and the fourth is
$$\frac{1}{(a-1)^2}
\left(a + a^{\lfloor \log_a n \rfloor}
(\lfloor \log_a n \rfloor (a-1) -a)\right).$$
This bound represented by these four terms plus the leading term is
actually attained and cannot be improved upon. For the asymptotics we
only need the dominant term, which is
$$\left(a - \frac{1}{2} a \right)
\lfloor \log_a n \rfloor^2
a^{\lfloor \log_a n \rfloor}
= \frac{1}{2} a \lfloor \log_a n \rfloor^2
a^{\lfloor \log_a n \rfloor}.$$
Now for the lower bound, which occurs with a one digit followed by
zeroes to give
$$T(n) \ge a^{\lfloor \log_a n \rfloor}
+ \sum_{j=0}^{\lfloor \log_a n \rfloor -1}
a^j \times (\lfloor \log_a n \rfloor - j) \times
a^{\lfloor \log_a n \rfloor-j}.$$
This simplifies to
$$a^{\lfloor \log_a n \rfloor}
+ \sum_{j=0}^{\lfloor \log_a n \rfloor -1}
(\lfloor \log_a n \rfloor - j) \times
a^{\lfloor \log_a n \rfloor}$$
which is
$$a^{\lfloor \log_a n \rfloor}
+ a^{\lfloor \log_a n \rfloor}
\sum_{j=0}^{\lfloor \log_a n \rfloor -1}
(\lfloor \log_a n \rfloor - j)$$
which finally produces
$$a^{\lfloor \log_a n \rfloor}
+ a^{\lfloor \log_a n \rfloor}
\sum_{j=1}^{\lfloor \log_a n \rfloor} j$$
or
$$a^{\lfloor \log_a n \rfloor}
+ \frac{1}{2}
\lfloor \log_a n \rfloor (\lfloor \log_a n \rfloor +1)
a^{\lfloor \log_a n \rfloor}.$$
The dominant term here is
$$\frac{1}{2}
\lfloor \log_a n \rfloor^2 a^{\lfloor \log_a n \rfloor}.$$
Joining the dominant terms of the upper and the lower bound we obtain
the asymptotics
$$\color{#00A}{\lfloor \log_a n \rfloor^2 \times
a^{\lfloor \log_a n \rfloor}
\in \Theta\left((\log_a n)^2 \times a^{\log_a n}\right)
= \Theta\left((\log n)^2 \times n\right)}.$$
This MSE link has a series of similar calculations.
Notice $\Theta(N)$ is constant, so the inhomogeneous term, $f(n)$ in the theorem, is $O(n^0)$. The critical exponent is $\log_b a = 0$. This equals the power of $n$ in the inhomogeneous term, so you are in case 2.
In case 2, we write $f(n) = \Theta(n^0 \log^0 n)$. The result of the master theorem is $T(n) = \Theta(n^0 \log^1 n)$. Which is correct, you expect to get $\log_2 n$ copies of $\Theta(N)$.
Here's a potentially more useful way to approach this problem, that highlights more of the theory...
We don't actually have $T(n) = T(n/2) + \Theta(N)$ because that explicitly makes $T$ dependent on $n$ and $N$ (and two constants about which we will have more to say). So, really, we have
$$ T(n,N) = T(n/2,N) + \Theta(N) \text{.} $$
That is, for each choice of $N$, there is a recurrence in $n$. If $N$ is chosen, we can suppress "$N$" in the recursive function $T$, which is what was done in the problem statement, but it is worth remembering that each choice of $N$ gives a different recursion. For each choice of $N$, we should think of this as an infinite list of facts: \begin{align*}
T(2,N) &= T(1,N) + c(2,N) \text{,} \\
T(4,N) &= T(2,N) + c(4,N) \text{,} \\
T(6,N) &= T(3,N) + c(6,N) \text{,} \\
&\vdots
\end{align*}
We have some bounds on the $c$s and there are two forms these bounds can take.
- There are (universal) contants, $k_1 > 0$, $k_2 > 0$, and $n_0 > 0$ such that for all $n > n_0, $
$$ k_1 N \leq c(n,N) \leq k_2 N \text{.} $$
- For each choice of $N$, there are constants (where we do not suppress the dependence on $N$) $k_1(N) > 0$, $k_2(N) > 0$, and $n_0(N) > 0$ such that for all $n > n_0$
$$ k_1(N) N \leq c(n,N) \leq k_2(N) N \text{.} $$
Which of these two is the case, if it is established at all, is established prior to the portion of the problem statement that is recited in the Question. For our purposes, it doesn't actually matter which is the case, so we assume the latter. If the former is the case, then each of the functions $k_1$, $k_2$, and $n_0$ become constant functions.
So now we know: for each choice of $N$, we have an infinite list of recursion facts. Each of these has a $c(n,N)$ and those $c(n,N)$ are bounded below and bounded above by bounds that are independent of $n$. These bounds only depend on $N$. Once $N$ has been chose, the bounds on the $c(n,N)$ are fixed and have no dependence on $n$ -- every $c$ on the list has the same bounds. (It is a useful exercise to show that the $n_0$ in the definition of $\Theta$ used above can be reduced to $0$. The short version is that the finite beginning of the list with $n \leq n_0$ is automatically bounded by multiples of $N$ and then the infinite rest of the list is bounded by multiples of $N$. If you choose the looser of the lower bounds and the looser of the upper bounds, you have bounded the entire list by mutiples of $N$, so the whole sequence is $\Theta(N)$.)
Let's look at what happens when we don't choose $N$ but compute $T(8,N)$.
\begin{align*}
T(8,N) &= T(4,N) + c(8,N) \text{,} \\
&= T(2,N) + c(4,N) + c(8,N) \text{, and} \\
&= T(1,N) + c(2,N) + c(4,N) + c(8,N) \text{.} \\
\end{align*}
We know that
\begin{align*}
k_1(N) N &\leq c(2,N) \leq k_2(N) N \text{,} \\
k_1(N) N &\leq c(4,N) \leq k_2(N) N \text{, and } \\
k_1(N) N &\leq c(8,N) \leq k_2(N) N \text{.}
\end{align*}
From these,
$$ 3 k_1(N) N \leq c(2,N) + c(4,N) + c(8,N) \leq 3 k_2(N) N \text{.} $$
(This inequality is what we mean when we write "$3\Theta(N)$" -- the sum of three, likely different, bounded numbers.)
Notice that the bounds on $c$ are independent of $n$. Whether $n$ is big or small, $c$ has the same bounds, so $\Theta(N)$ is both notationally and actually independent of $n$. That means $\Theta(N) \in O(n^0)$ (relative to $n$).
When we choose an $N$ and apply the master theorem to $T(n)$ (with the dependence on $N$ suppressed because we have already chosen $N$), we treat $n$ as the independent variable. Consequently, $\Theta$ and $O$ are relative to $n$. As we have shown $\Theta(N)$ is independent of $n$ -- the upper and lower bounds implicit in "$\Theta(N)$" are independent of $n$. We may replace $\Theta(N)$ with either the constant $k_1(N)N$ or the constant $k_2(N)N$, whichever leads to the worst case, and apply the master theorem to this recursion with constant inhomogeneous term. The result is described in the very first part of this Answer.
Best Answer
If $T(n)=T(n/2)+3T(n/4)+n$, one first centers the recursion using $S(n)=T(n)+cn$, then $$ S(n)=S(n/2)-cn/2+3T(n/4)-3cn/4+n+cn=S(n/2)+3S(n/4), $$ if $c=4$. From now on, $S(n)=T(n)+4n$.
Second, assume that $S(k)\bowtie c k^a$ for $k=n/2$ and $k=3n/4$ with $a\gt1$, where $\bowtie$ is either $\leqslant$ or $\geqslant$. Then $S(n)\bowtie cn^a/2^a+c3n^a/4^a=(1/2^a+3/4^a)cn^a$. From now on, assume that $a$ solves $$ \frac1{2^a}+\frac3{4^a}=1. $$ Then the property $S(n)\bowtie cn^a$ is hereditary for every fixed $c$.
In particular, for some small enough $c$ and some large enough $C$, it holds that $$ cn^a-4n\leqslant T(n)\leqslant Cn^a-4n, $$ for every $n$. Since $a\gt1$ in the present case, the contribution of the linear part of the recursion wins hence all this is more than enough to prove that $$ T(n)=\Theta(n^a). $$ One thing is clear though, which is that the fact that $1/2$ and $1/4$ are inverses of powers of a unique prime, invoked in the OP, does not enter the picture. Nevertheless, to relate $a$ to the exponent $\rho_1$ in the OP, assume that $a=1+\log_2\rho$, then $$ \frac1{2^a}+\frac3{4^a}=\frac1{2\rho}+\frac3{4\rho^2}, $$ hence $\rho$ solves $4\rho^2=2\rho+3$, that is, $\rho=\frac14(1\pm\sqrt{13})$, as claimed in the OP.
Edit: About the paragraph justifying the bounty, I would check the obvious references, namely generatingfunctionology and Flajolet's books.