Plot in the same figure the two curves
$$y=x\ ,\qquad y=f(x):=3x^3-7x^2+5x\ .$$
You then will see that $f(x)=x$ for $x=0, \ 1,\ {4\over3}$. Furthermore $f'(0)=5$, $f'(1)=0$, and $f'\bigl({4\over3}\bigr)={7\over3}$. This shows that $1$ is an attracting fixpoint of $f$ whereas the other two fixpoints are repelling. But we can learn more from the picture: If $x_0=a<0$ or $x_0=a>{4\over3}$ then obviously $\lim_{n\to\infty} x_n=-\infty$, resp. $=+\infty$.
When $x_0=a\in \ \bigl]0,{4\over3}\bigr[\ $ then things are more complicated. The function $f$ has a local maximum at $x={5\over9}$, but $f\bigl({5\over9}\bigr)={275\over243}<{4\over3}$, and a local minimum at the fixpoint $1$. This implies that in any case $$f\Bigl(\bigl]0,{4\over3}\bigr[\Bigr)\subset\bigl]0,{4\over3}\bigr[\ ,$$
so that an initial point $x_0\in \bigl]0,{4\over3}\bigr[\ $ can never escape to one of $\pm\infty$.
Note that $f\bigl({1\over3}\bigr)=1$. When $x_n<{1\over3}$ then $x_n