Is it possible to treat $\partial$ as an operator? I recently discovered that I can treat $d$ as an operator, and $dz$ as a certain variable defined as the infinitesimal change of $z$ for any variable $z$, and it is always defined with proportion to other variables $\left(\text{e.g.}  \, \frac{dz}{dx} \right)$, and that's how we use it in integrating, and that's how I can get some sense into why the fundamental theorem of calculus is true.
For example $z=x^2$ we have that $dz=2x\,dx$
and if $f=x^2+y^2$, we have that $df=2x \,dx + 2y \, dy$ by  partial differentiation. Now, I noticed that
$$y=x^2\implies dy=2x \,dx\implies d^2y=d(2x \, dx) = 2x\,d^2x+2(dx)^2$$
Here, I treated $d$ as an operator and $dx$ and $x$ as two separate  variables, and I used $d^2y=d(dy)=d(2x \, dx)$ and I used partial differentiation, again assuming that $dx$ and $x$ are two separate variables. Similarly, we can have
$$f=x^2+y^2\implies df=2x\,dx+2y\,dy \implies d^2f = 2 (dx)^2 + 2x \,d^2 x + 2(dy)^2+2y\,d^2y$$
This is also consistent with $\displaystyle \frac{d^2f}{(dx)^2} = 2+2\left(\frac{dy}{dx}\right)^2+2y\cdot\frac{d^2y}{(dx)^2}$
Note that this is consistent with the second derivative of $y$, we have that $\displaystyle \frac{d^2y}{(dx)^2}=2$ and from our last conclusion
$$\frac{d^2y}{(dx)^2}=2x\,\frac{d^2x}{(dx)^2}+2\cdot\frac{(dx)^2}{(dx)^2}$$
and since $\displaystyle \frac{d^2x}{(dx)^2}=\frac{d^2}{(dx)^2}x=\frac{d}{dx}\left(\frac{d}{dx}x\right)=0$, the above equation simplifies to $\displaystyle \frac{d^2y}{(dx)^2}=2$, consistent with our previous definitions.
I know that this is not a rigorous definition of something existing in Math (or may be it is and I'm not aware), but I sometimes use this and assume it somehow exists in Math.
Now, I tried to do similar things with the operator $\partial$, and I found inconsistencies. Is there a way to define $\partial$ as an operator and $\partial z$ as a variable the same way, and treat them with the same level of comfort I had while I treated $d$ here?
I haven't taken real analysis yet, but I'd still appreciate it if someone posted an answer based on real analysis with some preliminary explanation.
 A: The reason why you get inconsistencies using $\partial$ is that in the fraction $\frac{\partial y}{\partial x}$ there is information about the numerator in the denominator, so, when you split the fraction, you lose information.
A more straightforward notation, which does allow splitting the fraction, is to notate on the partial differential which variables were allowed to change.  So, for instance, a better notation would be:
$$ \frac{\partial y}{\partial x} = \frac{\partial_x y}{dx}$$
The right-hand side is a more precise notation.  You can get at most of the truths of partial derivatives by saying, if you have a function $f()$ of variables $x$, $y$, and $z$, then:
$$ df = \partial_x f + \partial_y f + \partial_z f$$
In other words, the total infinitesimal change in $f$ (i.e., $df$) is equal to the sum of the infinitesimal changes due to each individual variable.
Once you see this, then the other rules are just basic algebraic manipulations from this.
Example: the chain rule.  Let's say we wanted to know $\frac{df}{dt}$.  we get this by dividing both sides by $dt$:
$$ \frac{df}{dt} = \frac{\partial_x f}{dt} + \frac{\partial_y f}{dt} + \frac{\partial_z f}{dt} $$
Now multiply each term on the right by a variation of "1" that matches the partial.
$$ \frac{df}{dt} = \frac{\partial_x f}{dt}\frac{dx}{dx} + \frac{\partial_y f}{dt}\frac{dy}{dy} + \frac{\partial_z f}{dt}\frac{dz}{dz} $$
Now, swap the denominators of the right-hand side.
$$ \frac{df}{dt} = \frac{\partial_x f}{dx}\frac{dx}{dt} + \frac{\partial_y f}{dy}\frac{dy}{dt} + \frac{\partial_z f}{dz}\frac{dz}{dt} $$
Written in the typical notation, this becomes:
$$ \frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} + \frac{\partial f}{\partial z}\frac{dz}{dt} $$
