All rational periodic points I am trying to find all rational periodic points of a polynomial. To specify: a periodic point is the point that satisfy $f^n(x)=x$. It is related to dynamical systems in fact. So the current codes that I used are following:
A.<z> = AffineSpace(QQ, 1)
f = DynamicalSystem_affine([2*z^3-3*z^2+1/2])
x=f.dynatomic_polynomial(2)
x.factor()

With this I can find its dynatomic polynomial and factorize it and find rational roots of this polynomial. So this roots corresponds to periodic point of the polynomial of given period. In particular dynatomic polynomial is the polynomial of the form
$$ϕ_{n,f}(x)=∏_{d|n}(f^d(x)−x)^{μ(n/d)}$$
n is period, f is your polynomial and μ is the mobius function.
But with this code I can find periods up to 8 because of memory limit. The other code that I used is
R.<x> = QQ[]
K.<i> = NumberField(xˆ2+1)
A.<z> = AffineSpace(K,1)
f = DynamicalSystem([zˆ2+i], domain=A)
f.orbit(A(0),4)

But in fact it doesn't fit my purposes.
I have codes that I can get limited information. For example checking up to a period is not advisable. If you know a little bit arithmetic dynamics, you can see what I mean. Silverman-Morton conjecture plays an important role here.
I am waiting for your answers. Thank you so much.
 A: Your polynomial has good reduction outside of 2, so for all primes $p\ge3$, any rational preperiodic point will have period $n=m_p\cdot r_p\cdot p^{e_p}$, where $m_p$ is its period modulo $p$ and $r_p$ divides $p-1$ and $e_p$ is fairly small. (One can say more about $r_p$ and $e_p$ to pin things down further.) But in any case, looking at, say, $p=3,5,7$ and determining the periods of the periodic points in $\mathbb F_p$, you'll get a lot of information about the possible periods of $\mathbb Q$-rational periodic points.
Another standard trick is to use the fact that the periodic points will look like $A/2^k$ for some integer $A$. So you can compute the dynatomic polynomial using coefficients in $\mathbb F_p$ for a few medium size primes. That will get you congruence classes for its roots. This doesn't help with the fact that $\Phi_{n,f}(x)$ has pretty high degree, but it does prevent the coefficients from blowing up.
ADDENDUM (fixed as per Will Sawin's comments): Working mod $3$, the orbits are $\{0,2\}$ and $\{1\}$, hence  $m_2=1$ or $2$, and $r_2$ divides $3-1=2$. So the period of any rational point has the form $2^k\cdot3^\ell$ for some $0\le k\le 2$. Next we look mod $5$, where the periodic orbits are $\{0,3\}$ and $\{2\}$, since $4$ is preperiodic, but not periodic. Hence $m_5$ is $1$ or $2$, and $r_7$ divides $5-1=4$, so the period of any rational point has the form $2^m\cdot7^n$ for some $0\le m\le 3$. Combining the mod $3$ information with the mod $5$ information, we find that every rational periodic point has period $1$, $2$, or $4$.
Now one can compute $f^4(x)-x$ and factor it explicitly using a standard package (I used PARI GP). It has $7$ factors in $\mathbb Q[x]$, of which only three are linear. These lead to the fixed point $-\frac12$ and the orbit of size $2$ given by $\left\{0,\frac12\right\}$. So those are the only rational periodic points of your polynomial, unless you prefer to work in $\mathbb P^1$, in which case the point at infinity is also a fixed point.
As an alternative to computing $f^4(x)$, we can note that the multipliers of the periodic points mod $3$ are
$$ (f^2)'(0) = f'(0)f'(2) = 0 \pmod3
\quad\text{and}\quad
f'(1) = 0 \pmod3. $$
This implies that their $r_3$ values are both $r_3=1$. (In general, $r_p$ is $1$ if the multiplier of the point is $0$, and otherwise it is the order of the multiplier in the multiplicative group $\mathbb F_p^*$.) This rules out period $4$, so every rational periodic point has period $1$ or $2$.
ADDENDUM #2: Your polynomial has additional interesting properties. For example, it is post-critically finite, since the critical points are $0$, $1$ and $\infty$, which have orbits
$$
0 \to \frac12 \to 0,\quad
1 \to -\frac12 \to -\frac12,
\quad\text{and}\quad \infty\to\infty.
$$
