Answer
It is an increasing function, so it has an inverse (it is one-to-one.)
$\displaystyle \frac{df^{-1}}{dx}=\frac{1}{9}x^{-2/3}$
Work Step by Step
$x_{1} \gt x_{2}\qquad $... the cube function is an increasing function,
$x_{1}^{3} \gt x_{2}^{3}\qquad $... multiply with 27
$ 27x_{1}^{3} \gt 27x_{2}^{3}\qquad$
$ f(x_{1}) \gt f( x_{2})\qquad$
So f is an increasing funtion.
By Exercise 49, it is one-to-one and has an inverse.
Theorem 1 provides a way to find a formula for $\displaystyle \frac{df^{-1}}{dx}$
$\displaystyle \frac{d}{dx}[f^{-1}(x)]=\frac{1}{f'[f^{-1}(x)]}$
We find $f^{-1}(x)\quad \left[\begin{array}{l}
y=27x^{3}\\
x=27y^{3}\\
y^{3}=x/27\\
y=(x/27)^{1/3}\\
y=x^{1/3}/3\\
f^{-1}(x)=x^{1/3}/3
\end{array}\right]$
We find$ f'(x).$
$f'(x)=27\cdot 3x^{2}=81x^{2}$
Apply the formula:
$\displaystyle \frac{d}{dx}[f^{-1}(x)]=\frac{1}{f'[x^{1/3}/3]}=\frac{1}{81\cdot[x^{1/3}/3]^{2}}=\frac{1}{9x^{2/3}}$
$\displaystyle \frac{df^{-1}}{dx}=\frac{1}{9}x^{-2/3}$
