$$
\begin{aligned}
& \text { }(3 x-4 y)(d x-3 d y)+(6 d x-4 d y)=0 \\
& \Rightarrow \frac{d y}{d x}=\frac{(3 x-4 y)+6}{3(3 x-4 y)+4}
\end{aligned}
$$
Let $3 x-4 y=t \Rightarrow 3-4 \frac{d y}{d x}=\frac{d t}{d x}$
$$
\begin{aligned}
& \Rightarrow \quad \frac{1}{4}\left[3-\frac{d t}{d x}\right]=\frac{t+6}{3 t+4} \Rightarrow \frac{d t}{d x}=3-\frac{4(t+6)}{(3 t+4)} \\
& \Rightarrow \quad \frac{d t}{d x}=\frac{9 t+12-4 t-24}{(3 t+4)} \\
& \Rightarrow \quad \frac{d t}{d x}=\frac{5 t-12}{3 t+4} \Rightarrow d x=\left(\frac{3 t+4}{5 t-12}\right) d t \\
& \Rightarrow \quad d x=\left[\frac{3}{5}\left(\frac{5 t-12}{5 t-12}\right)+\frac{56}{5(5 t-12)}\right] d t \\
& \Rightarrow \quad x=\frac{3}{5} \int \frac{5 t-12}{5 t-12} d t+\frac{56}{5} \int \frac{d t}{5 t-12} \\
& \Rightarrow \quad x=\frac{3}{5} t+\frac{56}{25} \log |5 t-12|+C \\
& \Rightarrow \quad x=\frac{3}{5}(3 x-4 y)+\frac{56}{25} \log |15 x-20 y+12|+C \\
& \Rightarrow \quad 5 x-15 y+14 \log |15 x-20 y+12|=C
\end{aligned}
$$