г) $$\frac{3x+y}{x^2 + xy} - \frac{x+3y}{y^2 + xy}$$.

г) $$\frac{3x+y}{x^2 + xy} - \frac{x+3y}{y^2 + xy} = \frac{3x+y}{x(x+y)} - \frac{x+3y}{y(y+x)} = \frac{y(3x+y) - x(x+3y)}{xy(x+y)} = \frac{3xy + y^2 - x^2 - 3xy}{xy(x+y)} = \frac{y^2 - x^2}{xy(x+y)} = \frac{(y-x)(y+x)}{xy(x+y)} = \frac{y-x}{xy}$$;

Ответ: $$\frac{y-x}{xy}$$