A variant to Newton's method

I must consider the iteration $\displaystyle x_{n+1}=x_n - \frac{f(x_n)}{f'(x_0)}$. I must find the constants $\displaystyle C$ and $\displaystyle s$ such that $\displaystyle e_{n+1}=Ce_n ^s$.

Where $\displaystyle e_n=x_n-r$ and $\displaystyle r$ is the root of f.

My attempt: $\displaystyle e_{n+1}=x_{n+1}-r=x_n-\frac{f(x_n)}{f'(x_0)}-r=e_n-\frac{f(x_n ) }{f' (x_0 )}$.

So I must rewrite $\displaystyle e_n-\frac{f(x_n ) }{f' (x_0 )}$ as $\displaystyle C e_n ^s$. Thus I think I must relate $\displaystyle e_n$ to $\displaystyle \frac{f(x_n)}{f(x_0)}$ but I don't know how to proceed.

Any idea?

Thank you.