We have given: $$m\frac{d^2x}{dt^2} + V'(x(t))= 0\tag1$$
And: $$E = \frac{m}{2}\left(\frac{dx}{dt}\right)^2 + V(x(t))\tag2$$ where $V(x(t))$ is a known derivable potential.
So first we have to prove that E is independent of time so $\dfrac{dE}{dt} = 0$. It is trivial: $$\dfrac{dE}{dt} = \frac{dx}{dt}[m(\frac{d^2x}{dt^2}) + V'(x(t))]=0$$
And also i get: $$x(t)=\sqrt{\frac{2}{m}} \int\sqrt{E-V(x(t))}dt$$ and I'm stuck here because $V$ is a function of $x$ and not of $t$. I tried finding an expression from eq$(1)$ for $\frac{dx}{dt}$ and replacing it in the eq$(2)$ but it only gets more complicated and I end up back where I started. Any help would be greatly appreciated.
Unfortunately life is not that easy. All you can do with equation (2) is the following: $$ \dot{x}^2 = \frac{E-V}{m/2} \implies \pm\int\sqrt{\frac{m/2}{E-V}}dx = \Delta{t} $$ And all you can do with the integral is:
a) hope it's analytically solvable and
b) hope the anti-derivative is invertible. If both check out(which sometimes they do, for example in simple spring induced motion), you can have the motion of the particle.
Here's some suggested reading : https://en.wikipedia.org/wiki/Elliptic_integral , it comes up in the solutions to the simple pendulum which has the exact sort of setup your question has.