Answer
$\dfrac{3y}{3-x}\div\dfrac{12xy}{x^{2}-9}=-\dfrac{x+3}{4x}$
Work Step by Step
$\dfrac{3y}{3-x}\div\dfrac{12xy}{x^{2}-9}$
Factor the denominator of the second fraction:
$\dfrac{3y}{3-x}\div\dfrac{12xy}{x^{2}-9}=\dfrac{3y}{3-x}\div\dfrac{12xy}{(x-3)(x+3)}=...$
Evaluate the division of the two rational expressions and simplify by removing the factors that appear both in the numerator and the denominator of the resulting expression. To eliminate $(3-x)$ and $(x-3)$, change the sign of the factor $(3-x)$ and also the sign of the fraction:
$..=\dfrac{3y(x-3)(x+3)}{12xy(3-x)}=-\dfrac{3(x-3)(x+3)}{12x(x-3)}=-\dfrac{3(x+3)}{12x}=...$
$...=-\dfrac{x+3}{4x}$
