$$\frac{3x+y}{y} \cdot \left( \frac{y}{x} - \frac{3y}{3x+y} \right) = \frac{3x+y}{y} \cdot \left( \frac{y(3x+y)-3xy}{x(3x+y)} \right) = $$
$$\frac{3x+y}{y} \cdot \frac{3xy+y^2-3xy}{x(3x+y)} = \frac{3x+y}{y} \cdot \frac{y^2}{x(3x+y)} = \frac{y}{x}$$
Ответ: $$\frac{y}{x}$$