\( \lim _ { x \rightarrow 0 ^ { + } } \frac { 3 x ^ { 2 } - 6 x } { 3 \sqrt[ 3 ] { ( x ^ { 3 } - 3 x ^ { 2 } ) ^ { 2 } } } = \)

To evaluate the limit \[ \lim_{x \rightarrow 0^+} \frac{3x^2 - 6x}{3\sqrt[3]{(x^3 - 3x^2)^2}}, \] we start by simplifying the expression. The numerator can be factored as: \[ 3x(x - 2) \] The denominator simplifies as follows. First, observe that as \(x\) approaches \(0^+\), \[ x^3 - 3x^2 = x^2(x - 3). \] Thus, \[ (x^3 - 3x^2)^2 = (x^2(x-3))^2 = x^4(x - 3)^2. \] Taking the cube root gives: \[ \sqrt[3]{(x^3 - 3x^2)^2} = \sqrt[3]{x^4(x - 3)^2} = x^{\frac{4}{3}} \sqrt[3]{(x - 3)^2}. \] Substituting these into our limit expression results in: \[ \lim_{x \rightarrow 0^+} \frac{3x(x - 2)}{3x^{\frac{4}{3}} \sqrt[3]{(x - 3)^2}}. \] Now, the \(3\) in numerator and denominator cancels out: \[ \lim_{x \rightarrow 0^+} \frac{x(x-2)}{x^{\frac{4}{3}} \sqrt[3]{(x - 3)^2}} = \lim_{x \rightarrow 0^+} \frac{x^{\frac{2}{3}}(x - 2)}{\sqrt[3]{(x - 3)^2}}. \] As \(x\) approaches \(0^+\), \(x - 2\) approaches \(-2\) and \(\sqrt[3]{(x - 3)^2}\) approaches \(\sqrt[3]{9}\). Therefore, we get: \[ \lim_{x \rightarrow 0^+} \frac{(0) \cdot (-2)}{\sqrt[3]{9}} = 0. \] Hence, the limit is \[ \boxed{0}. \]