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 $v=v(x)$?
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 $$ a = v \frac{dv}{dx} $$
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.
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.
