The solution curve of the differential equation $\log _e\left(\frac{d y}{d x}\right)=3 x+4 y$, passing through the origin is

$4 e^{3 x}+3 e^{-4 y}-7=0$

$\log \left(\frac{d y}{d x}\right) =3 x+4 y \Rightarrow \frac{d y}{d x}=e^{3 x+4 y} \Rightarrow \frac{d y}{d x}=e^{3 x} e^{4 y}$

so  $\int e^{-4 y} d y =\int e^{3 x} d x$

$\Rightarrow \frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}+c$  as it passes (0, 0)

$\Rightarrow \frac{e^{-0}}{-4}=\frac{e^0}{3}+c \Rightarrow c=-\left(\frac{1}{3}+\frac{1}{4}\right)=-\frac{7}{12}$

so  $\frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}-\frac{7}{12} \Rightarrow-3 e^{-4 y}=4 e^{3 x}-7$

$\Rightarrow 4 e^{3 x}+3 e^{-4 y}-7=0$