How to prove $\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$ How could we prove that
$$ \frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$$
I have reduced it the form 
$$\frac{4^{\ln(4)/\ln(3/4)}}{3^{\ln(3)/\ln(3/4)}}$$
I am not sure what to do next to get snappy solution. Any ideas?

This refers to the original question, which had the left hand side equal to $\frac{1}{2}$ instead of $\frac{1}{12}$.
They are not equal.
Your simplification is correct. Then we can rewrite the left hand side as
$$\left(\frac{4^{\ln(4)}}{3^{\ln(3)}}\right)^{1/\ln(3/4)}$$
so raising both sides of the equation to the $\ln(3/4)$ power, we get that the equation would be equivalent to
$$\frac{4^{\ln(4)}}{3^{\ln(3)}} \stackrel{?}{=} \frac{1}{2}^{\ln(3/4)}.$$
Rewriting $4^{\ln(4)}$ as $e^{(\ln 4)^2}$, $3^{\ln(3)}$ as $e^{(\ln(3))^2}$, and $\left(\frac{1}{2}\right)^{\ln(3/4)}$ as $e^{-\ln(2)\ln(3/4)}$, the equality would be equivalent to
$$\left(\ln 4\right)^2 - \left(\ln 3\right)^2 \stackrel{?}{=} -\ln(2)\ln\frac{3}{4}.$$
Now, $\ln(4) = 2\ln(2)$, and $\ln\frac{3}{4} = \ln 3 - 2\ln 2$. So the left hand side is equal to
$$4(\ln 2)^2 - (\ln 3)^2$$
while the right hand side is equal to
$$-\ln(2)(\ln 3 - 2\ln 2) = 2(\ln 2)^2 - (\ln 2)(\ln 3).$$
But $$4(\ln 2)^2 - (\ln 3)^2 \approx 0.714863$$
and
$$2(\ln 2)^2 - (\ln 2)(\ln 3) \approx 0.199406$$

As corrected, the right hand side now be, after the simplification
$$
\left(\frac{1}{12}\right)^{\ln(3/4)} = \exp\left(-\ln(12)\ln(3/4)\right).$$
The exponent can be simplified:
$$\begin{align*}
-\ln(12)\ln(3/4) &= -\left(\ln(3)+2\ln(2)\right)\left(\ln(3)-2\ln(2)\right)\\
&= \left(\ln(3)+2\ln(2)\right)\left(2\ln(2)-\ln(3)\right)\\
&= \left(2\ln(2)\right)^2 - \left(\ln 3\right)^2\\
&= 4(\ln 2)^2 - (\ln 3)^2.
\end{align*}$$
Since this is the same as the exponent of $e$ on the left hand side, we do indeed have
$$\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}.$$

There's nothing special about $3$ and $4$. Replacing them with arbitrary positive numbers $a$ and $b$ will lead to 
$$\exp\left((\ln(a))^2 - (\ln(b))^2\right) \stackrel{?}{=} \exp\left(-\ln(ab)(\ln(b/a)\right)$$
which of course holds, since
$$-\ln(ab)\ln(b/a) = (\ln a + \ln b)(\ln a - \ln b)$$
giving the equality you have in the comment:
$$\frac{a^{1/\log_a(b/a)}}{b^{1/\log_b(b/a)}} = \frac{1}{ab}.$$
