Phương pháp giải:

Đồ thị hàm số \(y = f\left( x \right)\) nhận \(x = {x_0}\) làm TCĐ \( \Leftrightarrow \mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = \infty \).

Lời giải chi tiết:

Ta có:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{x}{{\sqrt {{x^3} + mx + 1}  - \sqrt[3]{{{x^4} + x + 1}} + {m^2}x}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{\dfrac{{\sqrt {{x^3} + mx + 1}  - \sqrt[3]{{{x^4} + x + 1}} + {m^2}x}}{x}}}\end{array}\)

Để đồ thị hàm số có TCĐ \(x = 0 \Rightarrow \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \infty  \Leftrightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^3} + mx + 1}  - \sqrt[3]{{{x^4} + x + 1}} + {m^2}x}}{x} = 0\).

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^3} + mx + 1}  - \sqrt[3]{{{x^4} + x + 1}} + {m^2}x}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^3} + mx + 1}  - \sqrt[3]{{{x^4} + x + 1}} + {m^2}x}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt {{x^3} + mx + 1}  - \sqrt[3]{{{x^4} + x + 1}}}}{x} + {m^2}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^3} + mx + 1}  - 1 - \sqrt[3]{{{x^4} + x + 1}} + 1}}{x} + {m^2}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^3} + mx + 1}  - 1}}{x} - \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[3]{{{x^4} + x + 1}} - 1}}{x} + {m^2}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {{x^3} + mx + 1}  - 1} \right)\left( {\sqrt {{x^3} + mx + 1}  + 1} \right)}}{{x\left( {\sqrt {{x^3} + mx + 1}  + 1} \right)}} - \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt[3]{{{x^4} + x + 1}} - 1} \right)\left( {{{\sqrt[3]{{{x^4} + x + 1}}}^2} + \sqrt[3]{{{x^4} + x + 1}} + 1} \right)}}{{x\left( {{{\sqrt[3]{{{x^4} + x + 1}}}^2} + \sqrt[3]{{{x^4} + x + 1}} + 1} \right)}} + {m^2}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^3} + mx + 1 - 1}}{{x\left( {\sqrt {{x^3} + mx + 1}  + 1} \right)}} - \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^4} + x + 1 - 1}}{{x\left( {{{\sqrt[3]{{{x^4} + x + 1}}}^2} + \sqrt[3]{{{x^4} + x + 1}} + 1} \right)}} + {m^2}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} + m}}{{\sqrt {{x^3} + mx + 1}  + 1}} - \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^3} + 1}}{{{{\sqrt[3]{{{x^4} + x + 1}}}^2} + \sqrt[3]{{{x^4} + x + 1}} + 1}} + {m^2}\\ = \dfrac{m}{2} - \dfrac{1}{3} + {m^2} = 0 \Leftrightarrow 6{m^2} + 3m - 2 = 0\,\,\left( * \right)\end{array}\)

(*) có 2 nghiệm trái dấu và có tích các nghiệm là: \({m_1}{m_2} = \dfrac{{ - 1}}{3}\).

Chọn D.