3. 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 $\qquad$ (where $[x]$ denotes the greatest integer not exceeding $x$).
(Let $\frac{3 y}{2 x}=t$, then the condition becomes $1+t=\left[\frac{y^{2}}{x^{2}}\right]$, so $1+t=\left[\frac{4 t^{2}}{9}\right] \Rightarrow 1+t \leqslant \frac{4 t^{2}}{9} \leqslant t+2$. Thus, $t=-1$ or $t=3$. Therefore, $\frac{y}{x}=-\frac{2}{3}$ or 2.)