= Solution
Write $\xi=V_1$ and $\eta=\phi$. The vector field
$$
V=\xi(x,u)\partial_x+\eta(x,u)\partial_u
$$
generates a <Lie point symmetry of an ordinary differential equation> if its local flow maps solution graphs to solution graphs. Its $n$th <prolongation of a vector field> is
$$
\operatorname{pr}^{(n)}V
=V+\sum_{j=1}^n\eta^{(j)}\partial_{u^{(j)}},
$$
where
$$
\eta^{(0)}=\eta,
\qquad
\eta^{(j)}=D_x\eta^{(j-1)}-u^{(j)}D_x\xi
$$
and
$$
D_x=\partial_x+u'\partial_u+u''\partial_{u'}+\cdots
$$
is the <total derivative operator>. The infinitesimal invariance criterion is
$$
\boxed{
\operatorname{pr}^{(n)}V(\Delta)=0
\quad\text{whenever }\Delta=0
}.
$$
Put $p=u'$ and $q=u''$. Direct calculation gives
$$
\eta^{(1)}
=\eta_x+(\eta_u-\xi_x)p-\xi_up^2
$$
and
$$
\eta^{(2)}
=\eta_{xx}+(2\eta_{xu}-\xi_{xx})p
+(\eta_{uu}-2\xi_{xu})p^2-\xi_{uu}p^3
+(\eta_u-2\xi_x-3\xi_up)q.
$$
For
$$
\Delta=q-\frac{p^2}{u}+u^2,
$$
the invariance condition is
$$
\eta^{(2)}
-\frac{2p}{u}\eta^{(1)}
+\frac{p^2}{u^2}\eta
+2u\eta=0
$$
after imposing $q=p^2/u-u^2$. The coefficient of $p^3$ is
$$
-\xi_{uu}-\frac1u\xi_u.
$$
It must vanish identically, so
$$
\partial_u(u\xi_u)=0.
$$
Since $u>0$, integration gives
$$
\boxed{\xi(x,u)=F(x)\log u+G(x)}
$$
for functions $F,G$.
Now take
$$
\xi=cx+d,
\qquad
\eta=-2cu.
$$
Then
$$
\eta^{(1)}=-3cp,
\qquad
\eta^{(2)}=-4cq,
$$
and substitution gives
$$
\operatorname{pr}^{(2)}V(\Delta)
=-4c\left(q-\frac{p^2}{u}+u^2\right)
=-4c\Delta.
$$
Thus the infinitesimal invariance criterion holds.
The generated one-parameter transformation is, for $c\ne0$,
$$
\boxed{
\widetilde x=e^{c\varepsilon}\left(x+\frac dc\right)-\frac dc,
\qquad
\widetilde u=e^{-2c\varepsilon}u
},
$$
while for $c=0$ it is $\widetilde x=x+d\varepsilon$, $\widetilde u=u$. Combining these transformations gives the two-parameter symmetry group
$$
\boxed{
(x,u)\longmapsto(\lambda x+a,\lambda^{-2}u),
\qquad \lambda>0,\ a\in\mathbb R
}.
$$
This is the <affine-scaling symmetry of u double prime equals u prime squared over u minus u squared>.
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