= Solution
Let a one-parameter infinitesimal transformation be $\delta q_i=\epsilon X_i(q,t)$ and suppose its change in the <Lagrangian> is a total derivative,
$$
\delta\mathcal L=\epsilon\frac{dF}{dt}.
$$
With <canonical momentum> $p_i=\partial\mathcal L/\partial\dot q_i$, the chain rule and the <Euler-Lagrange equation>[Euler-Lagrange equations] give, along a motion,
$$
\delta\mathcal L
=\epsilon\sum_i\left(
\frac{\partial\mathcal L}{\partial q_i}X_i
+p_i\frac{dX_i}{dt}\right)
=\epsilon\frac d{dt}\sum_i p_iX_i.
$$
Comparison with the assumed total derivative proves <Noether theorem>:
$$
\boxed{J=\sum_i p_iX_i-F\quad\text{is conserved}.}
$$
The given Lagrangian has no explicit time dependence, so its <energy function> is conserved:
$$
\boxed{E=\dot x\frac{\partial\mathcal L}{\partial\dot x}
+\dot y\frac{\partial\mathcal L}{\partial\dot y}-\mathcal L
=\frac{\dot x^2+\dot y^2}{2y^2}+V\!\left(\frac xy\right).}
$$
It is also invariant under the dilation $(x,y)\mapsto(e^\epsilon x,e^\epsilon y)$: the velocities and $y$ all scale by the same factor, while $x/y$ is unchanged. The generator is $(X_x,X_y)=(x,y)$ and $F=0$. Since
$$
p_x=\frac{\dot x}{y^2},
\qquad
p_y=\frac{\dot y}{y^2},
$$
Noether's theorem supplies the second first integral
$$
\boxed{D=xp_x+yp_y=\frac{x\dot x+y\dot y}{y^2}.}
$$
The integrals $E$ and $D$ are independent on an open dense subset of phase space.
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