Show that the iterative method $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}-\frac{f''(x_n)}{2f'(x_n)}\bigg(\frac{f(x_n)}{f'(x_n)}\bigg)^2$$
with $x_0$ given, used to approximate $s$ such that $f(s) = 0$, is of at least order $3$, provided that $f'(s) \neq 0$ where $s$ is a root.
Im really unsure how to answer this question so any help will be appreciated.
Let $h=-f'(x)^{-1}f(x)$ be the Newton update and compute the Taylor series of a shifted argument with unknown quadratic factor (using $O(h^k)=O(f(x)^k)$) \begin{align} f(x+h+ah^2)&=f(x)+f'(x)(h+ah^2)+\frac12f''(x)(h+ah^2)^2+O(h^3) \\ &=[f(x)+f'(x)h]+(af'(x)+\frac12f''(x))h^2+O(f(x)^3) \end{align} Now the first term vanishes by the definition of $h$ as Newton step, and the second is zero if $a=-\frac12f'(x)^{-1}f''(x)$, so that only third and higher order terms remain. This is the core for a proof of third order convergence.