= Solution
Put $q=u/x^2$, so $F=xG(q)$ and
$$
F_x=G(q)-2qG'(q),
\qquad
F_u=\frac1xG'(q).
$$
Substitution into the <Lie-symmetry determining equation for a first-order ordinary differential equation> and separation with respect to the arbitrary function $G$ gives
$$
\xi_u=0,\qquad
2q\xi-\frac\eta x=0,\qquad
\eta_x=0,\qquad
x(\eta_u-\xi_x)-\xi=0.
$$
These equations imply $\xi=Cx$ and $\eta=2Cu$. Apart from the zero generator, the common symmetry for arbitrary $G$ is therefore generated by
$$
\boxed{V=x\partial_x+2u\partial_u.}
$$
Its <one-parameter group> is the <scaling symmetry>
$$
\boxed{(x,u)\longmapsto(e^s x,e^{2s}u),}
$$
which leaves $u/x^2$ invariant. Special choices of $G$ can have additional symmetries, but this is the complete symmetry valid for an arbitrary smooth $G$.
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