I'd like to point out here that the expression «$\vec{a}=d\vec{v}/dt=\vec{v} d\vec{v}/dx$» is not entirely senseless.
In the definition
$$ \vec{a} = \frac{d\vec{v}}{dt} $$
velocity is considered to be the function of time $\vec{v} = \vec{v}(t)$ because we follow an individual particle along its trajectory. So in this case the expression is wrong.
But if you look at a continuous system of particles e.g. 359
flowing water, you can define the velocity to be a function of time and coordinates $\vec{v} = \vec{v}(t, x, y, z)$ - it would be equal to the velocity of a small droplet of water that happened to be in point $(x,y,z)$ at time $t$. As time goes by, this droplet would move to a different point in space $(x',y',z')$ , while another droplet would take its place at $(x,y,z)$ and so on. The $\vec{v}(t,x,y,z)$ is called "velocity field".
To calculate the acceleration of an individual droplet in this case, you should compare droplet's velocity at time $t$
$$ \vec{v}_d(t) = \vec{v}(t,x,y,z) $$
with the velocity it will have at time $t+dt$ having moved in space by $\vec{v}_d(t)dt$:
$$ \vec{v}_d(t+dt) = \vec{v}\left(t+dt, x+v_x(t,x,y,z)dt, y+v_y(t,x,y,z)dt, z+v_z(t,x,y,z)dt\right) $$
Expanding this you get
$$ \vec{v}_d(t+dt) = \vec{v}_d(t) + \vec{a}dt = \vec{v} + \left(\frac{\partial\vec{v}}{\partial t} + v_x \frac{\partial\vec{v}}{\partial x} + v_y \frac{\partial\vec{v}}{\partial y} + v_z \frac{\partial\vec{v}}{\partial z}\right)dt $$
so finally the acceleration is
$$ \vec{a} = \frac{\partial\vec{v}}{\partial t} + v_x \frac{\partial\vec{v}}{\partial x} + v_y \frac{\partial\vec{v}}{\partial y} + v_z \frac{\partial\vec{v}}{\partial z} $$ which is close to your original expression, but rather be written as
$$ \vec{a} = \frac{\partial\vec{v}}{\partial t} + (\vec{v} \cdot \frac{\partial}{\partial \vec{r}})\vec{v} $$ 766