Answer
$$ f^{-1}(x) =\sqrt[3]{\frac{1}{x}} $$
Domain of $ f^{-1}$ is $(-\infty,0) \cup (0,\infty)$
Range of $ f^{-1}$ is $(-\infty,0) \cup (0,\infty)$
Work Step by Step
Given $$ f(x) =\frac{1}{x^{3}}$$
Find the inverse:
$$y =\frac{1}{x^{3}} \\ x^{3} =\frac{1}{y} \\ x = \sqrt[3]{\frac{1}{y}} \\
\text {Switch } x \ and\ y, \text{}\\
y = \sqrt[3]{\frac{1}{x}} =f^{-1}(x)$$
We see that:
Domain of $ f^{-1}$ is $(-\infty,0) \cup (0,\infty)$
and
Range of $ f^{-1}$ is $(-\infty,0) \cup (0,\infty)$
Confirm the inverse:
\begin{aligned} f\left(f^{-1}(x)\right) &=f\left(\frac{1}{\sqrt[3]{x}}\right) \\ &=\frac{1}{\left(\frac{1}{\sqrt[3]{x}}\right)^{3}} \\ &=\frac{1}{1/x} \\ &=x \\f^{-1}(f(x)) &=f^{-1}\left(\frac{1}{x^{3}}\right) \\ &=\frac{1}{\sqrt[3]{\frac{1}{x^{3}}}} \\ &=\frac{1}{\frac{1}{x}} \\ &=x \end{aligned}
