Showing the invariance of the equations of motion It is strange to me that for a symmetry which involves $\dot{x}$, there seems to always appear a term with $\dddot{x}$ in the variation of the equations of motion, which doesn't makes much sense. I think that probably the procedure I am following is wrong.
I will show you an example: Consider the simple case of a free particle in one dimension, it's Lagrangian is:
$$L=\frac{1}{2}\dot{x}^2$$
It is obvious that the system conserves energy, so the symmetry that must be valid is $(\delta x=0 ,\delta t=\epsilon)$. I may rewrite these as symmetries that doesn't involve time variations (as the paper by E. L. Hill do):
$$(\delta_{*}x=\epsilon\dot{x},\delta_{*}t=0)$$
Now, I calculate the variation of the equations of motion, with the hope of finding that $\delta(\text{e.o.m})=\text{e.o.m}.$ Such result would mean that the equations of motion are invariant under the symmetry in consideration. So:
$$\delta{(\text{e.o.m})}=\delta(\ddot{x})=\ddot{\eta}$$
In this case $\eta = \dot{x}$ (remember that a variation is of the form $\delta{x}=\epsilon\eta$). So $\ddot{\eta}=\dddot{x}$. Hence:
$$\delta(\text{e.o.m})=\dddot{x}\neq\text{e.o.m}$$
This doesn't make sense because, for the system in consideration, the time translation is a Noetherian symmetry that gives conservation of Energy.

My question is: What is failing in this procedure? Is there a general way of showing that some Symmetry is indeed Noetherian?
 A: Short answer: In the reduction (when showing that an equation of motion $EOM=0$ has a symmetry), one is also allowed to use consequences of $EOM=0$. E.g. in OP's example:
$$\ddot{x}~=~ 0 \quad \Rightarrow \quad\dddot{x}~=~ 0,$$
See also e.g. this related Phys.SE post.
A: I'm not sure precisely what you did to solve this problem (for instance, how is the equation of motion, which is by definition an equation, also the quantity $ \ddot{x} $), but here's how I would approach it.
If we vary the Lagrangian according to the transformation you provide:
\begin{align}
\delta L & = \delta \left ( \frac{1}{2} \dot{x}^2 \right ) \\
& = \frac{1}{2} \delta \left ( \dot{x}^2 \right ) \\
& = \frac{1}{2} \cdot \frac{\partial \dot{x}^2}{\partial \dot{x}} \delta \dot{x} \\
& = \frac{1}{2} 2 \dot{x} \delta \dot{x} \\
& = \frac{1}{2} 2 \dot{x} \delta \frac{\partial x}{\partial t} \\
& = \frac{1}{2} 2 \dot{x} \frac{\partial}{\partial t} \left (  \delta x \right ) \\
& = \frac{1}{2} 2 \dot{x} \frac{\partial}{\partial t} \left (  \epsilon \dot{x} \right ) \\
& =  \dot{x} \frac{\partial}{\partial t} \left (  \epsilon \dot{x} \right ) \\
& = \frac{\epsilon}{2} \left [ \dot{x} \frac{\partial \dot{x}}{\partial t} + \frac{\partial \dot{x}}{\partial t} \dot{x} \right ] \\
& = \epsilon \frac{\partial}{\partial t} \left ( \frac{1}{2} \dot{x}^2 \right )
\end{align}
We now see $ L \to L + \epsilon \partial_{t} J $ for $ J = \frac{1}{2} \dot{x}^2 = L $. You can show that this always works out. Since the Lagrangian changed by only a total derivative, we have a symmetry, and you can show that the conserved charge is energy.
