Suppose $\epsilon > 0$ and that $c \in \mathbb{R}$. Since $f^3$ is continuous, we can choose $\delta > 0$ such that whenever $\lvert x - c \rvert < \delta$ we have that $\rvert f^3(x) - f^3(c) \lvert < \epsilon$.
To show that $f$ is continuous, I know I need some number $\delta_1 > 0$ such that $\lvert x - c \rvert < \delta_1$ implies that $\lvert f(x) - f(c) \rvert < \epsilon$.
I've tried manipulating the expression $\lvert f(x) - f(c) \rvert < \epsilon$ by using the fact that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. I get $$ \lvert f(x) - f(c) \rvert < \frac{\lvert f^3(x) - f^3(c)\rvert}{\lvert f^2(x) + f(x)f(c) + f^2(c) \rvert} < \frac{\lvert f^3(x) - f^3(c)\rvert}{\lvert f^2(x) + f(x)f(c) \rvert}. $$ However, I am not sure how to continue and I am also not confident that this is a good way to approach the exercise.
I suppose that the best way is to use the fact that the function $x\mapsto\sqrt[3]x$ is continuous. So,$$f^3\text{ continuous}\implies\sqrt[3]{f^3}\text{ continuous}\iff f\text{ continuous.}$$