Made a small observation, and I'm looking for a possible explanation.
Suppose we have $f(x)$ and we define $g(x)$ as: $$g(x):=x-\frac{f(x)}{f'(x)}$$ Why are the roots of $g'(x)$ and $f(x)$ almost identical?
Almost because although some are numerically equivalent, all the roots are not the same.
Consider $f(x):=x^2+3 x-1$, then $g'(x)$ is $\frac{2 \left(x^2+3 x-1\right)}{(2 x+3)^2}$, both functions' roots are: $$\left\{x\to \frac{1}{2} \left(-\sqrt{13}-3\right)\right\},\left\{x\to \frac{1}{2} \left(\sqrt{13}-3\right)\right\}$$ Consider $f(x):=x^5-2 x^4+2 x^3-2 x^2-x+\frac{1}{2}$, then the real roots are: $$\{x\to -0.52325\},\{x\to 0.33225\},\{x\to 1.66996\}$$ The roots of $g'(x)$ are: $$\{x\to 0.756324\},\{x\to -0.52325\},\{x\to 0.33225\},\{x\to 1.66996\}$$
You have $g' = \frac{f f''}{f'^2}$, therefore $g'(x_0) = 0$ if:
In your first example, $f''(x) = 2$ has no zeros, and all zeros of $f$ are simple, therefore $g'$ and $f$ have the same zeros.
In your second example the zero $x \approx 0.756324$ of $g'$ is a zero of $f''$.
Remark: $g$ is not defined at points where $f'(x)=0$. If $f$ is analytic (i.e. equal to its Taylor series) then it can be shown that $g'$ approaches either infinity or a finite non-zero limit at points where $f'(x_0) = 0$.