Chọn phương án B.

Đặt \(\log_9x=\log_6y=\log_4(2x+y)=t\).
Khi đó:

• \(x=9^t\)
• \(y=6^t\)
• \(2x+y=4^t\) (*)

Từ đó ta có phương trình $$\begin{aligned}
&\,2\cdot9^t+6^t=4^t\\
\Leftrightarrow&\,2\dfrac{9^t}{4^t}+\dfrac{6^t}{4^t}=1\\
\Leftrightarrow&\,2\left(\dfrac{3}{2}\right)^{2t}+\left(\dfrac{3}{2}\right)^t=1\\
\Leftrightarrow&\,2\left(\dfrac{3}{2}\right)^{2t}+\left(\dfrac{3}{2}\right)^t-1=0\\
\Leftrightarrow&\left[\begin{array}{ll}
\left(\dfrac{3}{2}\right)^t=-1 &\text{(vô nghiệm)}\\
\left(\dfrac{3}{2}\right)^t=\dfrac{1}{2}
\end{array}\right.
\end{aligned}$$
Khi đó: \(\dfrac{x}{y}=\dfrac{9^t}{6^t}=\left(\dfrac{3}{2}\right)^t=\dfrac{1}{2}\).