6. If real numbers $x, y$ satisfy
$$
4 x+3 y-2 x\left[\frac{x^{2}+y^{2}}{x^{2}}\right]=0,
$$

then the value of $\frac{y}{x}$ is $(\quad$.
(A) $\frac{2}{3}$ or 2
(B) $-\frac{2}{3}$ or 2
(C) 2
(D) None of the above
6.B.

Obviously, $4 x+3 y-2 x\left[\frac{x^{2}+y^{2}}{x^{2}}\right]=0$ $\Leftrightarrow 2+\frac{3 y}{2 x}=\left[1+\frac{y^{2}}{x^{2}}\right]$. Then $\frac{3 y}{2 x} \in \mathbf{Z}$. Let $\frac{3 y}{2 x}=t$. Thus, $2+t=\left[1+\frac{4 t^{2}}{9}\right]$ $\Rightarrow \frac{4 t^{2}}{9}<2+t \leqslant 1+\frac{4 t^{2}}{9}$
$\Rightarrow t=-1$ or 3 ( $t$ is an integer) $\Rightarrow \frac{y}{x}=-\frac{2}{3}$ or 2.