Is this an undergraduate-level proof of conservation of energy, or an arbitrary manipulation of symbols that happens to give the right answer? This is a slightly farcical question, for which I apologise.
An undergraduate tutee of mine was faced with the following problem:
Q. A particle of mass $m$ moving along a line is subject to a force $F(x) = −dV /dx.$ Show that the energy $E =\frac{1}{2}mv^2+V(x)$ is constant.
Here is his answer.
A. We're given $F(x)=-dV/dx.$ Now $F=ma$, so $m\ddot{x}=-dV/dx .$
Next observe $\ddot{x}=\frac{d\dot{x}}{dt}=\frac{d\dot{x}}{dx}\frac{dx}{dt}=\frac{d\dot{x}}{dx}\dot{x}$.
Hence $m\dot{x}\frac{d\dot{x}}{dx} = -dV/dx$, and integrating wrt $x$ we get $\frac12m\dot{x}^2 + V(x)=C$ a constant.
Is that a proof? On some formal level it looks fine to me. However an applied mathematician friend of mine (I'm a pure mathematiaian) objected that the UG had only proved that the energy was constant independent of $x$, rather than independent of $t$. I have no idea what that even means, on some level. On the other hand whilst I'm completely happy with the formal aspects of the argument, I have this uncomfortable feeling that if I were pressed to define $\frac{d\dot{x}}{dx}$ I would go for something like "change in $\dot{x}$ divided by change in $x$ and then take the limit" and if then someone observed that if the particle were stationary and $V$ were constant then $x$ wouldn't actually be changing so what does it even mean to change it, I think I'd be beginning to feel uncomfortable. The bottom line was that the marker didn't like it, because they claimed that something like $d(\dot{x})/dx$ was meaningless, and didn't give him any marks. I want to argue that usually it's meaningful but on the other hand...what am I saying here. I think I'm saying that part of me wants to resort to arguing that this proof is supposed to be applying to some space of potentials $V$, which are supposed to be at the very least continuous, so perhaps we could attempt to put some extra structure on the space of all potentials (e.g. restrict to $x$ in some closed and bounded interval and put some $L^2$ norm on it or something) and the potentials for which the argument seems a bit vague are some small subset of this space and so by continuity we should be fine. On the other hand this incident simply brings flooding back all the memories of the struggles I had doing applied mathematics when I was an undergraduate myself, when I would just try and answer the questions by formally manipulating the symbols and hoping that what I did would be bought by the people marking the questions. 
Presumably there are people out there that have a sufficiently rigorous way of thinking about this sort of question that they can say for sure whether this argument actually deserves any marks? Is it really an argument that "isn't quite rigorous but can be made rigorous using a standard technique"? Or is it fine? Or, heaven forfend, is it a question for which different people might have different opinions??
 A: I know this question is a little old(but could prove useful later) but a proof I used whilst completing my undergrad in the physics dept was
1) multiply both sides by the velocity
$$
m\ddot{x}\dot{x}=-\frac{dV}{dx}\dot{x}
$$
Then pull out the time derivative to yield
$$
\frac{d}{dt}\frac{m\dot{x}^2}{2}=-\frac{dV}{dt}
$$
The rhs is just reverse of the chain rule.
Pulling everything to the left hand side, and subbing in v for the velocity we find 
$$
\frac{d}{dt}\left[\frac{mv^2}{2}+V(x)\right]=0
$$
This leads to the conclusion that the energy in the system(kinetic + potential) I.e the terms in the brackets are independent of time and therefore constant or conservative.
You can also use this trick in solving the linear pendulum problem.
A: This proof appears valid to me. The justification of $\ddot{x} = \dot{x} \tfrac{d\dot{x}}{dx}$ via chain rule is acceptable mathematically, and even though the equations involved are supposed to have physical meaning, there is no need to explain what $\tfrac{d\dot{x}}{dx}$ means. From a physical level, what we have shown is that Work, which is defined as the change in kinetic energy, can be expressed as
$$ W = \Delta K = \int F \, dx$$
Then, since $F = -dV/dx$, then we have precisely shown that $\Delta K = - \Delta V$, so the total mechanical energy $K + V$ is constant.
Concerning the discussion above: While the above formally works, the main issue is that $\tfrac{d\dot{x}}{dx}$ may not be well defined. The crux of the problem is that the quantity
$$W = \int m\ddot{x} \, dx(t)$$
is in fact a line integral along the path of $x(t)$. If $t_0$ and $t_{k+1}$ are the initial and final times, and $t_1, \ldots, t_k$ are times at $\dot{x}$ changes sign, then we can turn the above into a sum of ordinary integrals against a Stieltjes measure (e.g. the variable we're integrating against is monotone), and so we have
$$ W = \sum_{i=0}^k \int_{t_i}^{t_{i+1}} m \ddot{x} \, dx = \sum_{i=0}^k \int_{t_i}^{t_{i+1}} m \ddot{x} \frac{dx}{dt} \, dt = \int_{t_i}^{t_{i+1}} \frac{1}{2} m\frac{d}{dt} \left[\dot{x}^2 \right]\, dt = \Delta \left( \frac{1}{2} m \dot{x}^2\right)$$
A: The marker was absolutely wrong to argue that $\frac{d \dot{x}}{dx}$ is meaningless! It is the spacial rate of change of velocity, which is a perfectly well defined concept. If the velocity doesn't change at all then $\frac{d\dot{x}}{dx}$ is 0, for this to be true then there are no forces on the particle to accelerate it so the potential gradient must also be zero (so the potential must be constant through space). The equation for the potential gradient the student derived is correct and integration wrt x is perfectly valid for this one dimensional problem, and proves the conservation of energy of a particle moving through space. Since the particle energy is conserved at each space point it arrives at, the energy must also be conserved through time.In the problem for motion in more than one dimension, integration wrt the space dimensions makes little sense-I think this is why some folk think the student was misguided in his approach. Eg consider motion in 2 dimensions: If we have a potential V(x,y) then the force $F_x$ in the x direction is $$F_x= - \frac{\partial{V(x,y)}}{\partial{x}}.$$ Similarly for the y direction and $$dV = \frac{\partial{V}}{\partial{x}} dx + \frac{\partial{V}}{\partial{y}} dy .$$ so $$- \frac{d{V}}{d{t}}= m (\frac{d^2x}{dt^2}.\frac{dx}{dt} + \frac{d^2y}{dt^2}.\frac{dy}{dt})= m(v_x \frac{dv_x}{dt}+v_y \frac{dv_y}{dt}) .$$ So integrate wrt time gives $$-V(x.y)=\frac{mv_x^2}{2}+\frac{mv_y^2}{2} + const.$$ So (Total KE + PE) = constant in time. It is also constant at each point in space along the path of the motion.
