Found Lagrange Equation $$E = 2F\ddot q + \frac{dF}{dq} + \frac{dv}{dq} $$
knowing that Equation for Total energy is E = T + v $$E = 2F\ddot q + \frac{dF}{dq} - \frac{dv}{dq} $$ Is that correct? ( change of signs? ) How would i find Time derivative $$ \frac{DE}{dt}? $$
Thanks in advice!
I have tried to decipher what you mean. It seems you are working with a Lagrangian
$$L(q,\dot{q}) = T-V = F(q) \dot{q}^2 - V(q)$$
The Euler-Lagrange equations coming from this Lagrangian, $\frac{dL}{dq} - \frac{d}{dt}\frac{dL}{d\dot{q}} = 0$, is
$$0 = 2F(q)\ddot{q} + \frac{dF(q)}{dq}\dot{q}^2 + \frac{dV(q)}{dq}$$
This is the same equation as you have, but in your question this equation is equated with the energy which is incorrect. The total energy is the sum of kinetic and potential energy
$$E = T + V = F(q) \dot{q}^2 + V(q)$$
To compute the time-derivative of the total energy $\frac{dE}{dt}$ use the chain-rule
$$\frac{dV(q)}{dt} = \frac{dV(q)}{dq} \frac{dq}{dt}$$
and similar for the other quantities. If you then use the Euler-Lagrange equation derived above you should be able to simplify the result and show that $\frac{dE}{dt} = 0$.