Can the idea of energy conservation be derived from Newtons's laws? From inspection of his laws you can vaguely discern a relationship but I want to know if you can manipulate his laws to prove it.
If not, what else did it take in history? What other assumptions about the world were needed to take that next step? Or is it only a fact of nature that can be proven experimentally, much like Newton's laws.
The answer is not. Newton's laws are equivalent to linear momentum conservation, but it does not implies mathematically energy conservation.
The elementary proof that is usually given assumes that involved forces derive from a potential energy. In that case the answer is yes, taking into account potential energy.
But consider for example two bodies of the same mass $m$ in rest at time $t=0$, which experiment a time dependent repulsive force $f(t)=mt, t \geq 0$ ($f_1(t)=mt$,$f_2(t)=-mt$). Energy at $t=0$ is $E=0$. The force verifies Newton's third law. Momentum is conserved because $p(t)=mt^2/2-mt^2/2=0$ but the kinetic energy is $E(t)=1/2mt^4/4+1/2mt^4/4$, which increases over time.
Newton's third law tells us that the momentum imparted on one body is equal and opposite to the momentum imparted on another if they interact. We then have $$ \Delta \vec p_1~=~-\Delta\vec p_2. $$ The change in momentum is $\Delta \vec p_i~=~m\vec a_i\Delta t$, $i~=~1,~2$. The change in momentum is with Newton's second law due to a force so that $$ \vec F_1~=~-\vec F_2. $$ The work on each body is $$ W_i~=~\int \vec F_i\cdot d\vec x. $$ However by $\vec F_1~+~\vec F_2~=~0$ it is clear that $W_1~+~W_2~=~0$ and the total energy is conserved.
No external forces conserves momentum and energy. If there is an external input of a force that is equivalent to the input of energy from outside the system. In that case to get energy conservation you have to then consider the nature of how this energy is input, or the reservoir of such energy.
Newton's second law, force f is $$f=m\frac{d^2 x}{d t^2}$$ x is position vector of the particle. $$f=-\frac{d v}{dx}$$v is the potential energy. $$m\frac{d^2 x}{d t^2}=-\frac{d v}{dx}$$ Multiply both sides with $\dot x$ $$\frac{m}{2} \frac{d\dot x^2}{dt}=-\frac{dv}{dt}$$ $$ \frac{d}{dt}(\frac{1}{2}m\dot x^2+v)=0$$ i.e., $$\frac{dE}{dt}=0$$Energy is conserved.
