Let $c \ne 0, a \in \mathbb{R}$ and let $x_0=c, \ x_{k+1}= \frac{2}{3}(x_k + \frac{a}{x_k^2})$ then compute the limit of the sequence and show the convergence.
I tried to solve the continuum problem associated to the sequence, unfortunately it didn't let me conclude anything.
Any suggestions?
Let me assume $a>0$. The other case should be pretty much symmetrical. Let $$ F(x)=\frac{2}{3}\left(x+\frac{a}{x^2}\right),\qquad x\neq 0$$ Simple algebraic manipulations show the following facts:
the only solution to $F(x)=x$ is given by $\bar{x}=(2a)^{1/3}$.
$F(x)>x$ for all $x\in J:=(-\infty,0)\cup (0,\bar{x})$.
$F(x)<x$ for all $x\in I:=(\bar{x},+\infty)=\left\{x:x>\bar{x}\right\}$.
We want to prove convergence to $\bar{x}$ for all $c\neq 0$. The case $c=\bar{x}$ is obvious.
First off, since $F$ is continuous on $\mathbb{R}\setminus \left\{0\right\}$, if the sequence $\left\{x_k\right\}$ converges, it must converge either to the only fixed point $\bar{x}$ or to $0$ (recall that $0$ does not belong to the domain of $F$). But the latter is actually impossible, because $\lim_{x\to 0}F(x)=+\infty$. So it suffices to prove that $\left\{x_k\right\}$ converges for all $c\neq 0$.
Now notice that $I$ is invariant under $F$, i.e. $F(I)\subset I$. Indeed, for all $x> \bar{x}$ (i.e. $x\in I$) we have $$F'(x)=\frac{2}{3}\left(1-\frac{2a}{x^3}\right)> \frac{2}{3}\left(1-\frac{2a}{\bar{x}^3}\right)=0$$ and hence $F$ is strictly increasing on $I$, so $x>\bar{x}$ (i.e. $x\in I$) implies $F(x)> F(\bar{x})=\bar{x}$, (i.e. $F(x)\in I$). Thus if $c>\bar{x}$, the sequence $\left\{x_k\right\}$ is decreasing (with $x_k> \bar{x}$ for all $k\in \mathbb{N}$), and hence it converges.
Finally, if $c\in J$, then if $x_k\in I$ for some $k\in \mathbb{N}$ we are done from what we proved above. Otherwise, we have $x_k\leq \bar{x}$ for all $k\in \mathbb{N}$, and since $F(x)>x$ for all $x\in J$, the sequence $\left\{x_k\right\}$ is increasing, and therefore convergent.