Numerical analysis fixed point iteration on $g(x) =x−af(x)−b(f(x))^2−c(f(x))^3$

I am not sure if I understand the following problem correctly nor if I am on the right path. Here it is, with my idea for a proof:

Consider the fixed point iteration method $$x_{k+1}=g(x_k)$$, $$k= 0,1,...$$ for solving the nonlinear equation $$f(x) = 0$$. Consider choosing an iteration function of the form:

$$g(x) =x−af(x)−b(f(x))^2−c(f(x))^3$$ where a, b and c are parameters to be determined. Find expressions for the parameters a, b and c such that the iteration method is of fourth order.

Idea: If $$r$$ is a root such that $$f(r)=0$$ then $$g'(r)=0$$, $$g''(r)=0$$, $$g'''(r)=0$$ but $$g^{IV}(r)\neq 0$$. As such, $$g'(r) =1−af'(r)−2b(f(r))f'(r)−3c(f(r))^2f'(r)=0$$ which means $$a=\frac{1}{f'(r)}$$.

Is this the right approach to find a? Where can I take it next? Thanks and regards,

The basic fixed-point iteration to find roots of $$f$$ has the form $$g(x)=x-a(x)f(x)\implies g'(r)=1-a(r)f'(r).$$ So to get $$g'(r)=0$$, one needs $$a(r)=1/f'(r)$$. As $$r$$ is unknown, one can only set $$a(x)=1/f'(x)+b(x)f(x)$$ to get the correct value at $$x=r$$ while retaining the freedom to further adapt the formula. In consequence $$g(x)=x-\frac{f(x)}{f'(x)}-b(x)f(x)^2\\ g'(x)=\frac{f(x)f''(x)}{f'(x)^2}-b'(x)f(x)^2-2b(x)f'(x)f(x)$$ This has a double root at $$x=r$$ if $$\frac{f''(r)}{f'(r)^2}-2b(r)f'(r)=0 \implies b(x)=\frac{f''(x)}{2f'(x)^3}+c(x)f(x).$$ Now do a similar computation to determine the form of $$c(x)$$.
One could alternatively start with a formula with known 4th convergence order, $$\tilde g(x)=x+p\frac{(1/f)^{(p-1)}}{(1/f)^{(p)}},~~~p=4-1=3,$$ treat the fraction as a division of Taylor series in $$f(x)$$ or $$s=f(x)/f'(x)$$ and carry it out to get a truncated Taylor series expansion for the quotient.
Another related approach is to directly employ the Taylor expansion of $$f$$, we want $$f(g(x))=O(f(x)^4)$$. So \begin{align} f(g(x))&=f(x)-f'(x)(af+bf^2+cf^3)+\tfrac12f''(x)(af+bf^2+cf^3)^2-\tfrac16f'''(x)(af+bf^2+cf^3)^3+O(f(x)^4) \\ &=(1-af'(x))f(x)+(-bf'(x)+\tfrac12a^2f''(x))f(x)^2 \\&\quad+(-cf'(x)+abf''(x)-\tfrac16a^3f'''(x))f(x)^3+O(f(x)^4) \end{align} also gives easy-to-solve equations if one wants the coefficients to be zero.