= Solution
Let $e_n=x_n-x_*$ for the <limit> $x_*$. <Quadratic convergence> means that eventually $|e_{n+1}|\leq C|e_n|^2$ for a fixed constant $C$; exact asymptotic order two has $|e_{n+1}|/|e_n|^2\to c$ with $0<c<\infty$.
For this iteration and $x_*=3$, direct algebra gives
$$
x_{n+1}-3=x_n-6+\frac9{x_n}=\frac{(x_n-3)^2}{x_n}.
$$
Using the stipulated convergence $x_n\to3$, the ratio of successive squared errors is
$$
\boxed{\frac{|e_{n+1}|}{|e_n|^2}=\frac1{|x_n|}\longrightarrow\frac13.}
$$
Thus the convergence is quadratic. The iteration is well defined here: $x_1=7/2$, and thereafter $x_n>3$ implies $3<x_{n+1}<x_n$. The initial point is not already the <limit>, so the error ratio is meaningful.
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