Let $f\in C^3(\mathbb{R})$, suppose there exists $r\in\mathbb{R}$ such that
$$f(r)=f'(r)=0,f''(r)\neq0$$
The iterations are
$$x_{n+1}=x_n-\frac{\alpha f(x_n)}{f'(x_n)}$$
with a starting point $x_0$ that was chosen such that $x_n$ converges to $r$.
Find an $\alpha$ for which the rate of convergence is quadratic.
Prove for this $\alpha$ that
$$\lim\limits_{n\to\infty}\frac{|x_{n+1}-r|}{|x_n-r|^2} = \frac{f'''(r)}{6f''(r)}.$$
Consider the (equivalent) fixed point iteration $x_{n+1} = g(x_n)$ with $g(r) = r$.
Lemma If $ g'(r) = \cdots = g^{(m-1)}(r) =0$ and $g^{(m)}(r) \neq 0$, then $x_{n+1} = g(x_n)$ converges to the root with order $m$ and asymptotic error constant $\frac{1}{m!} |g^{(m)}(r)|$.
Proof: Observe that \begin{align*} g(x_n) &= g(r) + (x_n -r) g'(r) + \cdots + \frac{(x_n - r)^{m-1}}{(m-1)!} g^{(m-1)}(r) + \frac{(x_n -r)^m}{m!} g^{(m)}(\xi_n) \\ &= g(r)+\frac{(x_n -r)^m}{m!} g^{(m)}(\xi_n) \\ &= r + \frac{(x_n -r)^m}{m!} g^{(m)}(\xi_n) \end{align*} Since $g(x_n) = x_{n+1}$, we have $\lim \limits_{n \to \infty} \dfrac{ |x_{n+1} -r |}{|x_n -r|^{m}} = \lim \limits_{n \to \infty} \dfrac{1}{m!} |g^{(m)}(\xi_n)| = \frac{1}{m!} |g^{(m)}(r)|.$ QED
Main Problem: Here, the problem is $x_{n+1} = g(x_n)$ where $g(x_n) = x_n - \alpha \frac{ f(x_n)}{f'(x_n)}$, with $f(r) = f'(r)=0$ and $f''(r) \neq 0$.
You should check that $g(r)=r$ and that $g'(r)=0 \iff \alpha =2$. Directly calculating $\frac{1}{2!}|g''(r)|$ will give you the same asymptotic error constant as described in the problem statement.
This is closely related to the modified Newton's method, as $r$ is a root of $f(\cdot)$ of multiplicity 2. See some of the related posts in the sidebar for further ideas.