= Solution
For $V_1=\partial_x$, one has $\xi=1$, $\eta=0$, and hence $D_x\xi=0$. The recursion from part a gives $\eta_j=0$ for every $j$.
For $V_2=x\partial_x$, one has $D_x\xi=1$. Starting from $\eta_0=0$, induction gives
$$
\eta_j=-ju_j.
$$
For $V_3=x^2\partial_x$, one has $D_x\xi=2x$. The first coefficients are $\eta_1=-2xu_1$ and $\eta_2=-2u_1-4xu_2$, and induction gives
$$
\eta_j=-2jx u_j-j(j-1)u_{j-1}.
$$
Thus the three <prolongations of the projective vector fields on the line> are
$$
\boxed{
\begin{aligned}
\operatorname{pr}^{(n)}V_1
&=\partial_x,\\
\operatorname{pr}^{(n)}V_2
&=x\partial_x-\sum_{j=1}^n j u_j\partial_{u_j},\\
\operatorname{pr}^{(n)}V_3
&=x^2\partial_x-\sum_{j=1}^n
\bigl(2jx u_j+j(j-1)u_{j-1}\bigr)\partial_{u_j}.
\end{aligned}}
$$
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