2. Given $2 x-3 \sqrt{x y}-2 y=0(x>0)$. Then the value of $\frac{x^{2}+4 x y-16 y^{2}}{2 x^{2}+x y-9 y^{2}}$ is ( ).
(A) $\frac{2}{3}$
(B) $\frac{4}{9}$
(C) $\frac{16}{25}$
(D) $\frac{16}{27}$
2.D.

From the original equation, we get $2\left(\frac{x}{y}\right)-3 \sqrt{\frac{x}{y}}-2=0$.
Let $\sqrt{\frac{x}{y}}=t$, then the equation transforms into $2 t^{2}-3 t-2=0$, which is $(2 t+1)(t-2)=0$.
Solving, we get $t_{1}=2, t_{2}=-\frac{1}{2}$ (discard). Hence $\frac{x}{y}=4$.
Substituting $x=4 y$ into the fraction, we get
$$
\frac{x^{2}+4 x y-16 y^{2}}{2 x^{2}+x y-9 y^{2}}=\frac{(16+16-16) y^{2}}{(32+4-9) y^{2}}=\frac{16}{27}
$$