【解析】\(y = x - 1\)

可得无水平渐近线、铅直渐近线，故求斜渐近线即可．

\(k = \lim\limits_{x \to \infty} \frac{\sqrt[3]{x^3 - 3x^2 + 1}}{x} = \lim\limits_{x \to \infty} \sqrt[3]{\frac{x^3 - 3x^2 + 1}{x^3}} = 1\)

\(\begin{align}b &= \lim\limits_{x \to \infty} \sqrt[3]{x^3 - 3x^2 + 1} - x \\&= \lim\limits_{x \to \infty} x \cdot \left( \left( 1 + \frac{1 - 3x^2}{x^3} \right)^{\frac{1}{3}} - 1 \right) \\&= \lim\limits_{x \to \infty} x \cdot \frac{1}{3} \cdot \frac{1 - 3x^2}{x^3} = -1\end{align}\)

故\(y = x - 1\)．