= Solution
Write the <Schwarzian derivative> as
$$
S=\frac{u_3}{u_1}-\frac32\frac{u_2^2}{u_1^2}.
$$
The first prolonged generator is $\partial_x$, and $S$ has no explicit dependence on $x$, so $\operatorname{pr}^{(3)}V_1(S)=0$. Under $\operatorname{pr}^{(3)}V_2$, each $u_j$ has weight $-j$. Both terms in $S$ therefore have weight $-2$, giving $\operatorname{pr}^{(3)}V_2(S)=-2S$. Substitution of
$$
\eta_1=-2xu_1,\qquad
\eta_2=-2u_1-4xu_2,\qquad
\eta_3=-6u_2-6xu_3
$$
for $V_3$ gives $\operatorname{pr}^{(3)}V_3(S)=-4xS$. Therefore the <projective transformation weights of the Schwarzian derivative> are
$$
\boxed{c_1=0,\qquad c_2=-2,\qquad c_3=-4x.}
$$
For an equation $\Delta=S-f(x)=0$, the <Lie point symmetry of an ordinary differential equation> criterion is
$$
\boxed{\operatorname{pr}^{(3)}V_i(\Delta)=0
\qquad\text{whenever }\Delta=0.}
$$
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