My physics teacher told us that we can’t divide vectors, that vector division has no physical meaning or significance. How about this: $$a = vdv/dx.$$ It says acceleration vector equals velocity (as a function of $x$) times $dv$ ‘divided’ by $dx$.
Here both $dv$ and $dx$ are vectors. How do I make sense of it? Because vector division doesn’t exist in physics right?
The statement $a = v (dv/dx)$ only holds in that form for one-dimensional motion, where the quantities $v$ and $x$ are just numbers rather than vectors. It follows from the chain rule, if we view $v$ as a function of $x$ instead of as a function of $t$: $$ a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{dv}{dx} v. $$
If you're doing 2-D or 3-D motion, you can still do something similar, but you have to let $\vec{v}$ be a function of $x$, $y$, and $z$, since $\vec{v}$ can change as each of these quantities changes. This means that you need to use multi-variable calculus to write out an equivalent statement. For example, we have $$ a_x = \frac{dv_x}{dt} = \frac{\partial v_x}{\partial x} \frac{dx}{dt} + \frac{\partial v_x}{\partial y} \frac{dy}{dt} + \frac{\partial v_x}{\partial z} \frac{dz}{dt} \\= \frac{\partial v_x}{\partial x} v_x + \frac{\partial v_x}{\partial y} v_y + \frac{\partial v_x}{\partial z} v_z. $$ As you can see, we're never "dividing by" the entire vector $\vec{x}$ when we take these derivatives; we only ever "divide by" its components $x$, $y$, or $z$, which is a valid mathematical operation.
Your expression doesn't make sense for (non-trivial) vectors, only for scalars (which are one-dimensional, a.k.a. "trivial" vectors).
The expression $a=v\frac{dv}{dx}$ only makes sense if $v$ and $x$ are both functions from one dimension to one dimension. You will never see the expression $\vec{a}=\vec{v}\frac{d\vec{v}}{d\vec{x}}$, because such an operation is not defined.
In general, multiplication of vectors is only well-defined in specific circumstances. There are three commonly-used multiplications of vectors: the dot product (which returns a scalar), the cross product (which returns a vector), and the tensor product (which returns a matrix). The cross product, in addition, only makes sense for three-dimensional and seven-dimensional vectors (the reason for this has to do with the existence of certain extensions of complex numbers called the quaternions and the octonions).
Division of two vectors is in general not defined, because it's not possible to undo either type of vector multiplication. For example, for any given vector, there are an infinite number of other vectors whose dot product with that vector will be zero (namely, all vectors that are perpendicular to it), and similarly, there are an infinite number of other vectors whose cross product with that vector will be zero (namely, all vectors that are parallel to it). Defining an inverse would require a one-to-one correspondence between the input and output of this multiplication, so defining division is not strictly possible.
That said, defining differentiation of vectors is different, namely because differentiation does not involve dividing two vectors. For a vector which is a function of space $\vec{v}(\vec{x})$, there are four ways to define differentiation:
$$\frac{d\vec{v}(\vec{x},t)}{dt}=\left\langle \frac{d v_1(\vec{x},t)}{d t},...,\frac{d v_n(\vec{x},t)}{d t}\right\rangle$$
$$\nabla \cdot \vec{v}(\vec{x})=\sum_{i=0}^n \frac{\partial v_i(\vec{x})}{\partial x_i}$$
$$\nabla \times\vec{v}(\vec{x})=\left\langle\frac{\partial v_y(\vec{x})}{\partial z}-\frac{\partial v_z(\vec{x})}{\partial y},\frac{\partial v_z(\vec{x})}{\partial x}-\frac{\partial v_x(\vec{x})}{\partial z},\frac{\partial v_x(\vec{x})}{\partial y}-\frac{\partial v_y(\vec{x})}{\partial x}\right\rangle$$
$$J_{ij}=\frac{\partial v_i(\vec{x})}{\partial x_j}$$
As you can see, none of these involve dividing a vector by a vector; in fact, all of them involve multiplying a vector by another vector, namely, the vector of partial derivative operators $\nabla$. For the divergence, this is a dot product; for the curl, this is a cross product; and for the Jacobian matrix, this is a tensor product.
So how do we rework the one-dimensional expression for acceleration so that it makes sense in any number of dimensions? The key is to start with the basic definition, assuming, as you did, that there is no explicit time dependence for $\vec{v}$:
\begin{align} \vec{a}(\vec{x})&=\frac{d\vec{v}(\vec{x})}{dt}\\ &=\left\langle \sum_{i=1}^n \frac{\partial v_1(\vec{x})}{\partial x_i}\frac{\partial x_i}{\partial t},...,\sum_{i=1}^n\frac{\partial v_n(\vec{x})}{\partial x_i}\frac{\partial x_i}{\partial t}\right\rangle\\ &=\left\langle \sum_{i=1}^n \frac{\partial v_1(\vec{x})}{\partial x_i}v_i,...,\sum_{i=1}^n\frac{\partial v_n(\vec{x})}{\partial x_i}v_i\right\rangle\\ &=(\vec{v}\cdot\nabla)\vec{v} \end{align}
There is nothing wrong with differentiating a vector, since it doesn't involve division by a vector.
Well, we know how to multiply vectors, via the dot product $\vec a \cdot \vec b$. So we could define a division through:
$$ \frac{\vec a}{\vec b} \doteqdot \vec a \cdot \frac{1}{\vec b}. $$
Now we need to define what is meant by $ \frac{1}{\vec b}$. Well, what about:
$$ \frac{1}{\vec b} \doteqdot \frac{\vec b}{|\vec b|^2}. $$
This vector inversion gives a new vector with same direction, but reciprocal magnitude. In fact it's how we define an inversion for other things like the special conformal transformation. Seems like as good a definition as any. Then division of vectors becomes:
$$ \frac{\vec a}{\vec b} = \frac{\vec a \cdot \vec b}{|\vec b|^2} = \frac{|a|}{|b|}\cos\theta. $$
This division satisfies what you would expect:
$$ \frac{\vec a}{\vec a} = 1, $$
but different from usual division, the inverse of $\vec a$ isn't unique.
First off, a key thing to know is that $\frac{dy}{dx}$ is not actually a division. It is the symbol representing taking the derivative of y with respect to x. The symbol looks like a division because that proved convenient. There's a remarkable number of cases where if you treat it as if it were a division, you get the right answer. However, it does not always work that way, so you can't actually think of it as a division.
As others have pointed out, in multivariable cases, we actually switch the symbol to $\frac{\partial y}{\partial x}$, reminding ourselves that the multi-variable case has a few extra things you have to track. You can't quite use the same old equations. We'll also use symbols liks $\nabla\vec v$ to capture concepts like that.
There is, however, a side conversation that I find very fascinating which is that of real division algebras. These are operations over fields of real numbers (where +, - and * do the operations you expect) where we can meaningfully define division. Mathematicians are very pedantic about this, so they define that to mean that for any A and non-zero B
One can see that any reasonable meaning of division will have to be layered on top of such a pattern.
So if we consider these real fields, like (x), (x, y), (x, y, z), (x, y, z, w), etc. we can try to find ways to define them such that those two division rules make sense. As it turns out, due to a rather clever little theorem, division can only work in 1, 2, 4, and 8 dimensional cases (or infinite dimensional... but that's another story entierly).
And that's it. Those are literally the only finite-dimensional real fields which can admit division.
This means your 3d vector not only doesn't have a division operator, but in fact there is no possible way 3 real numbers can be put together to create a meaningful division algebra. It actually cannot be done!
You can show that you can define a 2d vector division, but that it must be the same (isometric, technically) as complex division, so you can ask yourself whether it would make sense to think of your vector as a complex number or not. If it doesn't make sense, then there wont be a way to make division make sense either.
While I concur there is no way to divide vectors, generally speaking, it is fun to think of creative possibilities.
The concepts of inverses and uniqueness etc are studied in "abstract algebra", an area of mathematics often introduced at 2nd year university level. For example a structure with some sort of "addition" and "multiplication" is called a ring. A nicer type of ring where every non-zero element has an inverse is called a field. (This is very imprecise, I encourage people to look up the specific details.) Our first example is a field. Example 2 does not have a multiplicative identity.
A simple-minded way of seeing this is as follows: consider a linear function A that changes a vector x into another vector y: y = Ax. If you "divide by x", you get: A = y/x. So, we could say that the result of dividing vector y by vector x is a more complicated object, i.e., a tensor: the matrix A.
A natural way to define vector division is as the inverse of vector-scalar multiplication. Since you can multiply a scalar s by a vector v to obtain a vector w, where s * v = w
...then you could define w/v = s.
It's not earth-shattering, but could be useful in some contexts. It would have whatever physical interpretation that the inverse of vector-scalar multiplication would have.
