This is the problem I'm doing and the start of the explanation of how to do it:
Use Newton's Method to approximate to $4$ decimal place: $$\sqrt[3]{11}$$
This would be the equivalence of finding the zero for $x^3-14=0$. So that would make our function $f(x)=x^3-14$. $x_1=3$ will be our starting point since it is closest perfect cube root.
Based on other example problems, I thought it should be $x^3-11 = f(x)$. Trying that, using the formula $x_{n+1} = x_{1} - \frac{x^3-11}{3x^3}$ I just kept getting negative numbers that just bounced back and forth, eventually, between -1.something and -2.something. So with my understanding of how I get f(x) is wrong, or the example problem is wrong again I somehow either screwed up the overly simple math of setting the cube root of a 11 to zero, or (despite checking and adding plenty of parentheses) I'm failing at my calculator input. Or some combination of those. I need held figuring what.
It must have been some typo in the formula that you have used. If you do everything correctly you will get something like this