I have this distribution function
f(x) = \begin{cases} 1-{x^\frac{-1}3)}, & \text{if $x$ >=1 } \\ 0, & \text{if $x$ < 1} \end{cases}
to find the density function I know that I need to do an integral, but in some reason I dont get the correct answer,
this is the correct answer
f(x) = \begin{cases} \frac{1}3x^\frac{-4}3, & \text{if $x$ >=1 } \\ 0, & \text{if $x$ < 1} \end{cases}
If the cumulative distribution function is $F$, then the probability density function is $F'$. You seem to have the terms backwards.
The first function $F$ given by
$$F(x)=\begin{cases} 1-x^{-1/3} & \text{if $x \geq 1$ } \\ 0 & \text{if $x< 1$} \end{cases}$$
is the distribution function. (You can see this because $F$ is nondecreasing and continuous and $\lim_{x\to-\infty}F(x)=0$ and $\lim_{x\to\infty}F(x)=1$).
The second function $f$ given by $$f(x)=\begin{cases} \frac{1}{3}x^{-4/3} & \text{if $x\geq 1$ } \\ 0 & \text{if $x< 1$} \end{cases}$$
is the density function.
Given the cumulative distribution function $F$, find the density function $f$ by differentiation:
$$f(x)=F'(x)$$
for all $x<1$ and all $x>1$.
If instead you were given the density function $f$, you can find the cumulative distribution function $F$ using integration:
$$F(x)=\int_1^x f(t)dt$$
for all $x\geq 1$.