Let $f\in C^2$ and $x^*$ be a simple root of $f$, i.e. $f(x^*)=0\wedge f'(x^* )\ne 0$. Further, let $U(x^*):=\left\{x : |x-x^* |\le r\right\}$ for some $r>0$ and
$\;\;\;\;\;\;\;\;\;\;\displaystyle y^{(n)}:=x^{(n)}-\frac{f(x^{(n)})}{f'(x^{(n)})}\;,\;\;\;x^{(n+1)}:=y^{(n)}-\frac{f(y^{(n)})}{f'(x^{(n)})}$
describe the two-stage Newton method. I want to show that $\left(x^{(n)}\right)_{n\in\mathbb{N}_0}$ converges locally with order $3$ to $x^*$.
Proof $\;\;\;f\in C^1$ and $f'(x^*)\ne 0$ $\Rightarrow$ there exists a neighbourhood $U'(x^*)\subset U(x^*)$ such that
$\;\;\;\;\;\;\;\;\;\;|f'(x)|>0\;\;\;\forall x\in U'(x^*)$
Let $x^{(n+1)}\in U'(x^*)$. By using the mean-value theorem for differentiation, it holds
$\;\;\;\;\;\;\;\;\;\;\left|x^{(n+1)}-x^*\right|\le \left|y^{(n)}-x^*\right|\left|1-\frac{f'(\xi)}{f'(x^{(n)})}\right|$
for some $\xi\in (y^{(n)},x^*)$ and
$\;\;\;\;\;\;\;\;\;\;\left|1-\frac{f'(\xi)}{f'(x^{(n)})}\right|\le L_1\in (0,1)$
is satisfiable. Now, we can proceed in the same way with $y^{(n)}$ to obtain
$\;\;\;\;\;\;\;\;\;\;|y^{(n)}-x^*|\le \left|x^{(n)}-x^*\right|\left|1-\frac{f'(\zeta)}{f'(x^{(n)})}\right|$
for some $\zeta\in (x^{n)},x^*)$ and
$\;\;\;\;\;\;\;\;\;\;\left|1-\frac{f'(\zeta)}{f'(x^{(n)})}\right|\le L_2\in (0,1)$
is satisfiable, too. Taking everything into account we get
$\;\;\;\;\;\;\;\;\;\;|x^{(n+1)}-x^*|\le L_1|y^{(n)}-x^*|\le L_1L_2|x^{(n)}-x^*|$
However, while this is beautiful, it only leads to linear convergence. I've missed to use $f\in C^2$. What can I conclude from that and how will it help to show the convergence of order 3?
I could not figure the solution in - like half an hour to this. But you can improve your result to be at least of order 2, the same as the unmodified newton method. Consider your last line: $|y_n -x^*|$ $y_n$ is the simple newton method.
Expand the function $f$ into a Taylor series up to order 2 with lagrangian remainder at $x_k$
$f(x)=f(x_k)+f'(x_k)(x-x_k)+f''(\xi)(x-x_k)^2/2$
Now evaluate this at $x=x^*$ using the assumption that $f$ has a root there.
Then expand $|y_n-x^*|$ and solve the evaluated taylor expansion for an appropriate term.
Thus you will have shown that the method is at least of order 2.
I will look into the remainder of the task later today.