Can you remove the log?

Taking $log$ of both side of each equation and applying the properties $\log _{ c }(a^{r})= r\ log _{ c }(a)$ and $\log _{ c }(ab)=\log _{ c }(a)+\log _{ c }(b)$ we get $\log3 (\log 3+\log x)=\log 4 (\log 4 +\log y)$ and $\log4 \log x=\log 3 \log y$ Making $a=\log 3, b=\log 4, u=\log x, v=\log y$ we can obtain the following system of linear equations for $u$ and $v$ . $a(a+u)=b(b+v)\\ bu=av$ The only solution of the system is $u=-a$ and $v=-b$ . Thus $\log x =-\log 3 =\log \frac{1}{3}$ and $\log y =-\log 4 =\log \frac{1}{4}$ . Hence $x=\frac{1}{3}$ and $y=\frac{1}{4}$ and therefore $\frac{1}{x}+\frac{1}{y}=7$

$\begin{cases} (3x)^{\log{3}} = (4y)^{log{4}} \\ 4^{\log{x}} = 3^{\log{y}} \end{cases}$

$\Rightarrow \begin{cases} \log{3}\log{3} + \log{3}\log{x} = \log{4}\log{4} + \log{4}\log{y}\\ \log{x}\log{4} = \log{y}\log{3} \end{cases}$

$\Rightarrow \begin{cases} (\log{3})^2 + \log{3}\log{x} = (\log{4})^2 + \log{4}\log{y}\\ \log{y} = \dfrac {\log{4}\log{x}}{\log{3}} \end{cases}$

$\Rightarrow (\log{3})^2 + \log{3}\log{x} = (\log{4})^2 + \dfrac {(\log{4})^2\log{x}}{\log{3}}$

$\Rightarrow \left(\log{3} - \dfrac {(\log{4})^2}{\log{3}} \right) \log{x} = (\log{4})^2 - (\log{3})^2$

$\Rightarrow \left(\dfrac {(\log{3})^2 - (\log{4})^2}{\log{3}} \right) \log{x} = (\log{4})^2 - (\log{3})^2$

$\Rightarrow \log{x} = -\log{3}\quad \Rightarrow x = \frac {1}{3}$

Substituting $\space x = \frac {1}{3}\space$ in $\space (3x)^{\log{3}} = (4y)^{log{4}}$ :

$\Rightarrow 1 = (4y)^{log{4}} \quad \Rightarrow 4y = 1 \quad \Rightarrow y = \frac {1}{4}$

Therefore, $\space \dfrac {1}{x} + \dfrac {1}{y} = 3 + 4 = \boxed {7}$

just observe the equality and you will see it will occur when both sides are 1 in first equation and check that they satisfy eq 2 So x=1/3 and y=1/4