Now looking at this problem, one can simply tell that the values will be $\frac 13$ and $\frac 14$ for $x$ and $y$ respectively, since the terms will equate one. However, just out of curiosity, is there a way to actually solve it? I tried doing it, but I just can’t figure it out.
Thanks!
(For the base of the logs, just assume it as 10 and solve)
On
Use the shorthands $a = \log 3$, $b= \log4$ wnd take $\log(\cdot)$ on both sides to recast $(3x)^{\log 3}=(4y)^{\log 4}$ and $4^{\log x}=3^{\log y}$ respectively
$$a^2 +a\log x=b^2+b\log y\tag{1}$$
$$a\log y=b\log x\tag{2}$$
Then, (1)+(2) and (1)-(2) lead to
\begin{align} \log x+ \log y & =-a-b\\ \log x-\log y&=-a+b \end{align}
and, in turn, $\log x=-a,\space \log y = -b$, which yields the expected solutions
$$x=\frac{1}{3}, \space \space \space y=\frac{1}{4}$$
Taking the logarithm on both sides we get the first equation: $$\log(3)(\ln(3)+\ln(x))=\log(4)(\ln(4)+\ln(y))$$ doing the same with the second equation we obtain $$\ln(y)=\frac{\log(x)\ln(4)}{\ln(3)}$$ so we can substitute $\ln(y)$ in the first equation above to compute $x$. Assuming $\log=\ln$ we get $$\ln(3)^3-\ln(4)^2\ln(3)=\ln(x)(\ln(4)^2-\ln(3)^2)$$ so $$\ln(x)=\frac{\ln(3)^3-\ln(4)^2\ln(3)}{\ln(4)^2-\ln(3)^2}$$ The solution by MMA: $$\left\{\left\{x\to \exp \left(\frac{\log (3) \log (4)-\log ^3(3)}{\log ^2(3)-2 \log (2) \log (4)}\right),y\to \exp \left(\frac{2 \left(\log (2) \log (4)-\log (2) \log ^2(3)\right)}{\log ^2(3)-2 \log (2) \log (4)}\right)\right\}\right\}$$