So the function has $r = \sqrt[3]{R}$ as a zero for any positive real number R.
$$g(x) = 1/x^2 − x/R $$ Determine derivatives: $$g'(x)= -2/x^3 -1/R$$ $$g''(x)= 6/x^4$$
So I get for newtons iteration: $$x_{n+1} = 3Rx_{n}/(2R+ x_{n}^3)$$
Now this is some explanation i got but I dont follow.
Convergence analysis: consider $x > 0$ only. The function $g(x) $is monotone decreasing, convex diverging to $−∞$ when $x → ∞$. If $0 < x_{n} < r$, then $g(x_{n}) > 0$. Because of the decreasing convex properties, $x_{n} < x_{n+1} < r$, and from this $x_{n} → r$. Calculus shows that $x_{0} > r ⇒ 0 < x_{1} < r.$
Thus, convergence for all $x > 0.$
That the function $g$ is strictly convex implies that the tangents are all below the graph of the function. This in particular means that where the tangent $t_k$ at $x_k$, $$t_k(x)=g(x_k)+g'(x_k)(x-x_k)\le g(x),$$ has its root $x_{k+1}$, the function value is positive, $g(x_{k+1})>t_k(x_{k+1})=0$. So if you start at a point $x_0<r$ where the function $g$ is positive, then the Newton iterate $x_1$ will lie between this point and the root, giving monotonicity.
If $x_0>r>0$, then the Newton step formula shows that the tangent root $x_1=\frac{3Rx_0}{2R+x_0^3}$ is positive, from the convexity it follows that the function value $g(x_1)$ has to be positive, which implies $x_1<r$.