Answer
$y = 1$
Work Step by Step
Since
\begin{align*}
\lim _{t \rightarrow \infty} \frac{t^{1 / 3}-t^{-1 / 3}}{\left(t-t^{-1}\right)^{1 / 3}}&=\lim _{t \rightarrow \infty} \frac{1-t^{-2 / 3}}{\left(1-t^{-2}\right)^{1 / 3}}\\
&= \frac{1}{1^{1 / 3}}=1
\end{align*}
and
\begin{align*}
\lim _{t \rightarrow-\infty} \frac{t^{1 / 3}-t^{-1 / 3}}{\left(t-t^{-1}\right)^{1 / 3}}&=\lim _{t \rightarrow-\infty} \frac{1-t^{-2 / 3}}{\left(1-t^{-2}\right)^{1 / 3}}\\
&=\frac{1}{1^{1 / 3}}=1
\end{align*}
Then $f(x)$ has a horizontal asymptote of $y = 1$.
