Ta có \(y' = f'\left( x \right) \cdot g'\left( {f\left( x \right)} \right) = \left[ {4{x^3} + 2\left( {4 - {m^2}} \right)x} \right] \cdot g'\left( {f\left( x \right)} \right)\).

Yêu cầu bài toán \( \Leftrightarrow \left[ {4{x^3} + 2\left( {4 - {m^2}} \right)x} \right] \cdot g'\left( {f\left( x \right)} \right) \ge 0\,,\,\,\forall x \in \left( {0\,;\,\, + \infty } \right)\).

Lại có \(g'\left( x \right) = 3{x^2} - 6x + 5 = 3{\left( {x - 1} \right)^2} + 2 > 0\,,\,\,\forall x \in \left( {0\,;\,\, + \infty } \right)\).

Do đó \(g'\left( {f\left( x \right)} \right) > 0\,,\,\,\forall x \in \left( {0\,;\,\, + \infty } \right)\).

Suy ra \(4{x^3} + 2\left( {4 - {m^2}} \right)x \ge 0\,,\,\,\forall x \in \left( {0\,;\,\, + \infty } \right)\)

\( \Leftrightarrow 2x.\left( {{x^2} + 4 - {m^2}} \right) \ge 0 \Leftrightarrow {x^2} + 4 - {m^2} \ge 0\,,\,\,\forall x \in \left( {0\,;\,\, + \infty } \right)\)

\( \Leftrightarrow {m^2} \le {x^2} + 4\,,\,\,\forall x \in \left( {0\,;\,\, + \infty } \right) \Leftrightarrow {m^2} \le {\min _{\left( {0\,;\,\, + \infty } \right)}}\left( {{x^2} + 4} \right) = 4\).

Vậy \({m^2} \le 4 \Leftrightarrow - 2 \le m \le 2.\) Chọn C.