\(a{x^2}-ax-3x = -3\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,a{x^2}-\left( {a + 3} \right)x + 3 = 0.\)
Чтобы уравнение являлось квадратным должно выполняться условие: \(a \ne 0\).
\(D = {\left( {a + 3} \right)^2}-12a = {a^2} + 6a + 9-12a = {a^2}-6a + 9 = {\left( {a-3} \right)^2};\)
\({x_1} = \dfrac{{a + 3 + a-3}}{{2a}} = 1;\,\,\,\,\,\,\,\,\,{x_2} = \dfrac{{a + 3-a + 3}}{{2a}} = \dfrac{3}{a}.\)
\(\dfrac{{{x_1}}}{{{x_2}}} = \dfrac{1}{{\dfrac{3}{a}}} = 3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a = 9;\,\,\,\,\,\,\,\,\dfrac{{{x_2}}}{{{x_1}}} = \dfrac{{\dfrac{3}{a}}}{1} = 3\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,a = 1.\)
ОТВЕТ: 1; 9.