When can one write $a=v \cdot dv/dx$? Referring to unidimensional motion, it is obvious that it doesn't always make sense to write the speed as a function of position. Seems to me that this is a necessary condition to derive formulas like:
$$v^2=v_0 ^2 +2\int_{x_0}^{x}a\cdot dx$$
In fact, in the first step of the demonstration (the one I saw, but I think that this step is crucial) it's required to write $a=dv/dt=(dv/dx)(dx/dt)$, that doesn't make sense if $v$ isn't a function of $x$. 
When can one rigorously write $v=v(x)$?
 A: This is going to be essentially the same in content as Jerry Schirmer's response, but I thought you might like to hear it in more mathematical terms.  The velocity function $v$ is defined as
$$
  v(t) = \dot{ x}(t)
$$
Let's take the domain of the position function to be the open interval $(t_1, t_2)$ and suppose that it has the property that given any point $x_0$ in the range of $x$, there is a unique point $t_0$ in its domain $(t_1, t_2)$ such that $x(t_0) = x_0$.  Then there exists a function $x^{-1}$ (the inverse of $x$) defined on the range of $x$ satisfying
$$
  x^{-1}(x(t)) = t
$$
Now we define a function $\bar v$ on the range of $x$ by
$$
  \bar v(x) = v(x^{-1}(x))
$$
It is common to abuse notation here and use $v$ in place of $\bar v$ for this function, but let's keep things notationaly rigorous.  Then on one hand the chain rule gives
$$
  \frac{d}{dt}\bar v(x(t)) = \frac{d\bar v}{dx}(x(t))\,\dot x(t) = \frac{d\bar v}{dx}(x(t))\,v(t)
$$
While on the other hand we use the definition of $\bar v$ to write
$$
  \frac{d}{dt}\bar v(x(t)) = \frac{d}{dt} v(x^{-1}(x(t))) = \frac{dv}{dt}(t) = a(t)
$$
and combining these observations gives the identity you wanted
$$
  a(t) = \frac{d\bar v}{dx}(x(t))\,v(t)
$$
Notice that if we indulge in the usual abuse of notation, then we can simply write this as
$$
  a = v \frac{dv}{dx}
$$
A: It can be done in any case where the velocity can be written as a function of the position.  This can be done if the velocity is not constant, and if there are no turnaround points in the motion.  For instance, consider $x = r\sin(\omega t)$, $v = \omega r \cos(\omega t)$.  
Then, we have:
$$\begin{align}
x &= r \sin(\omega t)\\
t &= \frac{1}{\omega}\sin^{-1}(x/r)\\
v &= \omega r \cos (sin^{-1}(x/r))\\
&= \omega \sqrt{r^{2} - x^{2}}
\end{align}$$
Which is a valid transformation so long as $\sin^{-1}(x/r)$ is defined, which means that you only cover the right half of the unit circle.  
A: Generally, the total derivative can be broken into a sum of partial derivatives.  If the acceleration $a$ is taken to be a function only of $x$ and $t$, then the total derivative is
\begin{equation}
  a=\frac{\mathrm{d} v}{\mathrm{d} t} = \frac{\partial v}{\partial t} + \frac{\partial v}{\partial x}\frac{\partial x}{\partial t}
\end{equation}
One can safely write $\mathbf{a}=\mathbf{v}\cdot\nabla \mathbf v$, then, when $\partial v /\partial t=0$.  This is true when you are considering a single particle or object, as the velocity of the particle at a point where the particle doesn't exist is not changing.  However, for distributions of particles, the distinction is meaningful.
