If $\xi$ is a root of multiplicity $m$ of $f(x)=0$ and $x_0$ is close enough to $\alpha$ then the order of convergence of the sequence $\left\{x_n\right\}_{n \ge 0}$
defined by $$x_{n+1}=x_n-m\frac{f(x_n)}{f'(x_n)}$$
is at least $2$.
The sequence is a special case of functional iteration defined by $x_{n+1}=g(x_n)$ with $$g(x)=x-m\frac{f(x)}{f'(x)}$$
On the other hand I know that if $\xi$ is a fixed point of a function $g$ then if $$g^{(m)}(\xi)=0, \; m \in \left\{1,2,…,k-1 \right\} \;\;,\;\; g^{(k)}(\xi) \ne 0$$
Then the the functional iteration is convergent to $\xi$ assuming that $x_0$ is sufficiently close to $\xi$ and the order of convergence is exactly $k$.
So here the order of convergence is at least $2$ if the first derivative of $g$ at $x=\xi$ is zero, but:
$$g'\left(x\right)=1-m\frac{\left(f'\left(x\right)\right)^{2}-\left(f^{''}\left(x\right)\right)^{2}f\left(x\right)}{\left(f'\left(x\right)\right)^{2}}$$
So: $$g'\left(\xi\right)=1-m\frac{\left(f'\left(\xi\right)\right)^{2}}{\left(f'\left(\xi\right)\right)^{2}}=1-m$$
Which is not zero unless $m=1$ which means if $\xi$ is a simple root of $f(x)$, so what to do?
Moreover I think that does not make sense to define multiplicity of a root of an equation,rather it's defined for functions.
Best Answer
If $\xi$ is a root with multiplicity $m$, then $f(x)=(x-\xi)^m \phi(x)$, with $\phi(\xi)\ne 0$. So, $$ g(x)=x- m \dfrac{(x-\xi)^m \phi(x)}{m(x-\xi)^{m-1}\phi(x)+(x-\xi)^m \phi'(x)}=x-\dfrac{m(x-\xi)\phi(x)}{m\phi(x)+(x-\xi)\phi'(x)} $$
which in fact implies that $g'(\xi)=0$.