Answer
\begin{align*}
x_1 & = 2.66666 \\
x_2& = 2.6190\\
x_3& =2.6180
\end{align*}
Work Step by Step
Given $$ f(x)=x^{2}-3 x+1, \quad x_{0}=3$$
Since
$$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$
Then
\begin{align*}
x_{n+1}&= x_n -\frac{x_n^{2}-3 x_n+1}{2x_n-3}\\
&= \frac{x_n^{2}-1}{2x_n-3}
\end{align*}
Hence
\begin{align*}
x_1 &= \frac{x_0^{2}-1}{2x_0-3}= \frac{(3)^{2}-1}{2(3)-3}= 2.66666 \\
x_2&=\frac{x_1^{2}-1}{2x_1-3}= \frac{(2.66666)^{2}-1}{2(2.66666)-3}= 2.6190\\
x_3&= \frac{x_1^{2}-1}{2x_1-3}= \frac{(2.6190)^{2}-1}{2(2.6190)-3}=2.6180
\end{align*}
