Let $X$ be an absolutely continuous random variable.
I have to find the probability density function of $Y=X^{3}$ in terms of the probability density function of $X$.
To solve it I used the basic formula:
$f_{Y}(y)=f_{X}(g^{-1}(y))\left | \frac{\mathrm{d} g^{-1}(y)}{\mathrm{d} y} \right |$
With $g(x)=x^{3}$.
The problem is that to use this formula I have to verify that the $g$ function is strictly increasing and that the derivative of $g^{-1}$ exists but $\sqrt[3]{y}$ is not derivable at $0$. I want to know if that affects the reasoning. And if I'm wrong, what would be the right way?
As Henry says in the comments, the issue with the single point $x=0$ does not matter.
In case of doubt, brute-force it:
$$F_Y(y)=P(Y\le y) = P( X^3 \le y)=P(X \le y^{1/3})=F_X(y^{1/3}) \tag 1$$
We've used the property $a^3 \le b \iff a \le b^{1/3}$ (work out the details if not sure).
Finally $$f_Y(y)= \frac{d F_Y(y)}{dy}=f_X(y^{1/3}) \, \frac{1}{3} \, y^{-2/3}$$
This is undefined at $y=0$ (depending on $f_X$, it could have a finite limit), but anyway a single point (more precisely, a set of measure zero) is inconsequential for defining a density probability function.