So I need to prove that for $f(x) = ((x - x^*)^2)h(x)$ assuming $f''(x) \neq 0$ and assuming $f(x)$ be twice continuously differentiable, that Newton's method converges linearly in this case. So $g(x) = x - f(x)/f'(x)$.
I have a rate of convergence theorem that says for the derivatives of $g(x)$ the first derivative that equal zero when $x =$ the fixed point is the order of the convergence so if ($x^*$ is the fixed point) $g'(x*) = 0$ and $g''(x^*) \neq 0$ then the order of convergence is quadratic.
My problem is that in this case $g(x)$ at the fixed point doesn't exist since $f'(x^*)$ = 0 so $g(x^*) = x - 0/0$. I have other questions I need to do that end up like this two with the fixed point method given to me and the $f(x)$ leading to a $0/0$ situation, what do I do?
Since your presumable don't know the root when you using Newton's method the iteration is performed as $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ so technically you never divide by $0$. However, if $f'(r)=0$ at the root ($f(r)=0$), it can be proved, that the Newton method will give a linear rate of convergence theorem, instead of regular quadratic rate.
In order to show this one have to find the smallest $p$ for which there is a finite number $c\ne0$ such that $$\frac{\left|x_{n+1}-r\right|}{\left|x_{n}-r\right|^p}=c,$$ where $r$ is the root of $f$, i.e. $f(r)=0$.
$$\frac{\left|x_{n+1}-r\right|}{\left|x_{n}-r\right|^p}= \frac{\left|x_{n}-\frac{f(x_n)}{f'(x_n)}-r\right|}{\left|x_{n}-r\right|^p}\underset{h=x_n-r}{=} \left|h^{1-p} \left(1- \frac{f(x_n)}{h f'(x_n)}\right) \right| $$ using few terms of Taylor expansion around $r$ (since $x_n=r+h$)of the numerator and denumerator. I think you will also have to assume that $f''(r)\ne0$ or so. The full thing you find here.