\(The\ domain\ (D):\\x^2-9
eq0\ \wedge\ x-3
eq0\ \wedge\ 4x-12
eq0\\x^2
eq9\ \wedge\ x
eq3\ \wedge\ 4x
eq12\\x
eq-3\ \wedge\ x
eq3\\\boxed{D:x\in\mathbb{R}-\{-3;\ 3\}}\\\\\dfrac{x}{x^2-9}-\dfrac{1}{x-3}=\dfrac{1}{4x-12}\\\\\dfrac{x}{x^2-3^2}-\dfrac{1}{x-3}=\dfrac{1}{4(x-3)}\ \ \ \ |use\ a^2-b^2=(a-b)(a+b)\\\\\dfrac{x}{(x-3)(x+3)}-\dfrac{1}{x-3}=\dfrac{1}{4(x-3)}\\\\\dfrac{x\cdot4}{(x-3)(x+3)\cdot4}-\dfrac{1(x+3)\cdot4}{(x-3)(x+3)\cdot4}=\dfrac{1(x+3)}{4(x-3)(x+3)}\\\\\dfrac{4x-4(x+3)}{4(x-3)(x+3)}=\dfrac{x+3}{4(x-3)(x+3)}\iff4x-4x-12=x+3\)
\(x+3=-12\ \ \ \ |subtract\ 3\ from\ both\ sides\\\\\boxed{x=-15}\in D\)
