Division by $dx$ in multi-variable calculus .... I am stuck on this doubt :
Suppose $f=f(x,y,z).$ Hence, $ df= \frac {\partial f}{\partial x}dx + \frac { \partial f}{\partial y}dy + \frac {\partial f}{\partial z}dz.$ 
Then, is the following equation correct :
$$\frac {df}{dx}=\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}\frac{dy}{dx} + \frac {\partial f}{\partial z}\frac{dz}{dx} \,\,\,\,(*)$$
The reasoning used in obtaining $(*)$ is : "dividing" the whole equation by $dx$. Normally, the $\large \frac {d}{dx}$ operator is used in single variable calculus where only single-variable functions are differentiated wrt $x$. But it does look a bit awkward (at least to me) when used in multi-variable calculus. Do the expressions $\large \frac {df}{dx}$, $\large \frac {dy}{dx}$ and $\large \frac {dz}{dx}$ even make any sense when used like this ?
I know that "division" by $\partial x$ can cause problems in multi-variable calculus. But what about "division" by $dx$. It works fine in single-variable calculus.
If $\large \frac {df}{dx}$ makes any sense, then does it mean the "total" rate of change of $f$ wrt $x$ if $y$ and $z$ are allowed to change ?
Summary : 
(1) Is division by $dx$ allowed in multi-variable calculus? 
(2) What does $\frac {df(x,y,z)}{dx}$ mean if answer to (1) is "yes" ? Does it mean anything if the answer to $(1)$ is "no" ?
 A: For simplicity of notation, let $w$ represent $f(x,y,z)$ throughout.
Is division by $\mathrm dx$ allowed?
It depends on the context.
In calculus, notation like $\mathrm dx$ is usually something that you can't divide an arbitrary expression by. It's not a number, but a special notation whether a "differential form" or just part of the notation for derivatives of single variable functions, etc.
However, in some treatments of linear approximation (e.g. in Stewart's calculus texts), differentials like $\mathrm dx$ and $\mathrm dw$ gain a new meaning and division is quite reasonable. For independent/input variables like $x$, $\mathrm dx$ would just mean "a small change in the $x$ input", similarly to $\Delta x$; it would be a real number. For dependent/output variables like $w$, $\mathrm dw$ means something like "the linear approximation to the small change the output if the inputs are changed by $\mathrm dx$, etc."
What does $\dfrac{\mathrm df(x,y,z)}{\mathrm dx}$ mean if yes?
In the context of linear approximation where you would like to use those "number" meanings of the differentials: An equation like $\mathrm dw= \dfrac {\partial w}{\partial x}\mathrm dx + \dfrac { \partial w}{\partial y}\mathrm dy + \dfrac {\partial w}{\partial z}\mathrm dz$ is still true.
For a particular nonzero small change $\mathrm dx$, you could certainly divide the equation by it, to find that $$\dfrac{\mathrm dw}{\mathrm dx}= \dfrac {\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dx} + \dfrac { \partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dx} + \dfrac {\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dx}= \dfrac {\partial w}{\partial x} + \dfrac { \partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dx} + \dfrac {\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dx}$$
What does $\dfrac{\mathrm df(x,y,z)}{\mathrm dx}$ mean if no?
If we're not in a context where $\mathrm dx$ stands for a small real number, then the only meaning of $\dfrac{\mathrm dw}{\mathrm dx}$ we have access to is "the first derivative of $w$ with respect to $x$ (in a context where $w$ can be written as a function of $x$ alone)".
So again this depends on context. If $x,y,z$ are all independent variables, then I would say this notation is then meaningless. $\dfrac{\partial w}{\partial x}$ makes sense, but $w$ doesn't have a total derivative with respect to $x$ in this sense.
But if there is a dependence of $y$ and $z$ on $x$ in the context, then things are different. For instance, suppose we may write $y=g(x)$ and $z=h(x)$. Then as pointed out by Batominovski in a comment, we can make sense of $\dfrac{\mathrm dw}{\mathrm dx}$ even though division by $\mathrm dx$ would not be defined generally.
In that case, we can calculate $\dfrac{\mathrm df\left(x,g(x),h(x)\right)}{\mathrm dx}$ first by applying the multivariate chain rule to $f$ in the way you'd expect:
$$\dfrac{\mathrm df\left(x,g(x),h(x)\right)}{\mathrm dx}=\left.\dfrac {\partial f}{\partial x}\right|_{(x,y,z)=\left(x,g(x),h(x)\right)} + \left.\dfrac {\partial f}{\partial y}\right|_{(x,y,z)=\left(x,g(x),h(x)\right)}\dfrac{\mathrm dy}{\mathrm dx} + \left.\dfrac {\partial f}{\partial z}\right|_{(x,y,z)=\left(x,g(x),h(x)\right)}\dfrac{\mathrm dz}{\mathrm dx}$$
Especially in a physics text, it would not be too surprising to see the above equation abbreviated along the lines of $\dfrac {\mathrm dw}{\mathrm dx}=\dfrac {\partial w}{\partial x} + \dfrac { \partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dx} + \dfrac {\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dx}$.
A: Yes, it's correct, and what you've just computed is the rate of change of $f$ with respect to $x,$ where $y,z$ are also considered to be functions of $x.$
A: If $$f=f(x,y,z)$$ where $x,y,z$ are independent variables, then you may divide $df$ by $dx$.
For example $$f(x,y,z)= xyz+x^2+y^2+z^2$$
$$df = (yz+2x)dx + (xz+2y)dy + (xy+2z)dz$$
$$\frac {df}{dx} = yz+2x + (xz+2y)\frac {dy}{dx} + (xy+2z)\frac {dz}{dx} =yz+2x $$ 
Which is  the same thing as $\frac {\partial{f}}{\partial {x}}$
