The question is simple, find the $f(x)$ that was used for this fixed point iteration knowing it was created using Newton's Method:
$$x_{k+1} = 2x_k - x_k^2y$$
So we just rearrange to get it into $x - f(x)/f'(x)$ form:
$$ x_{k+1} = 2x_k - x_k^2y \\\\ x_{k+1} = x_k - (x_k^2y - x_k) \\\\ $$
So, $f(x)/f'(x) = x^2y - x$, however, I am unsure how to proceed further to find what $f(x)$ exactly is. Any help/direction would be appreciated.
It looks to me like $y$ is a constant in your question. I'll assume that in this answer.
You reached the point $f(x)/f'(x) = x^2y - x$. From there you can use $\frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}$. Plugging that into your equation, I get $$\begin{align} \frac{d}{dx} \ln(f(x)) &= \frac{1}{x^2 y - x} \\ \ln(f(x)) &= \int \frac{1}{x^2 y - t} dx \\ \ln(f(x)) &= \ln\left(\frac 1 x - y\right) + c \\ f(x) &= e^{\ln\left(\frac 1 x - y\right) + c} \\ f(x) &= e^c \left(\frac 1 x - y\right) \\ f(x) &= d \left(\frac 1 x - y\right) \\ \end{align}$$
Here $c$ and $d$ are arbitrary constants.
You can always double check the answer: we get $f'(x) = - \frac d {x^2}$, so $$\frac{f(x)}{f'(x)} = \frac{d \left(\frac 1 x - y \right)}{-d \frac 1 {x^2}} = -x^2 \left( \frac 1 x - y \right) = x^2 y - x$$ which was the relationship you wanted.