The solution of the differential equation $\frac{d y}{d x}=\frac{3 e^{2 x}+3 e^{4 x}}{e^x+e^{-x}}$

$y=\left(e^x+e^{-x}\right)+C$ $y=Ce^x$ $y=e^{3 x}+C$ $y=3 e^{3 x}+C$

$y=e^{3 x}+C$

The correct answer is Option (3) → $y=e^{3 x}+C$