Simple differential equation( introduction but need some basic explanation) I have a couple of questions before I dig deeper into my calculus book. 
First: 
I have learned that 
$\frac{d}{dx}\frac{x}{y}$=$\frac{y x'-x y'}{y^2}$
never really gotten a proper explanation for why suddenly when its differential equations it becomes as they have defined it in my book: 
$$\mathrm {d}(\frac{x}{y})=\frac{y\mathrm dx-x\mathrm dy}{y^2}$$
I see algebraically that this is simply dividing by dx, but I HATE when I do not get a proper explanation for it. 
Second is a more specific case from the start of the book: 
$$x^2 y\mathrm {dy}-(x\mathrm dy-y\mathrm dx)=0$$
Now If I divide by $x^2$
$$y \mathrm dy-\mathrm{d}\left[\frac{x}{y}\right]=0$$,
Now I get what the point of doing this is: 
However it is the result that throws me off: 
$y^2/2-x/y=c$ But I am integrating both why is the constant positive. Is it simply just that c can take on a negative value, why does not the c's cancel out since I integrate both. I am not sure how to interpret this. 
Last question:
$$\pm \frac{\mathrm {d}\left(x^2+y^2\right)}{2 \sqrt{x^2+y^2}}=\mathrm {dx}$$
Solving this produces 
$$\pm \sqrt{x^2+y^2}=c+x$$,
However, I do not see how this makes it any easier unless $\mathrm {d}\left(x^2+y^2\right)=2x$, but if that's the case I am even more confused
since in this case y disappears as I am used to dealing with derivatives since nowhere it is stated that $y$ is a function of $x$. But that to me conflicts what happens in what i wrote under First.
Could someone please explain in detail why in the last instance: $\frac{\mathrm d\left(x^2+y^2\right)}{\mathrm dx}=2 y \frac{\mathrm dy}{\mathrm dx}+2 x$ is not the case, but simply $2x$. In one moment I can use the derivative as a fraction(FIRST) in the other (last question) I can not?
 A: "dy/dx" is NOT a fraction but is defined as the limit of a fraction, the "difference quotient".  A result of that is that the derivative can often be treated like a fraction.  For example, if y is a function of u and u is itself a function of x, then y can be considered a function of x and, by the "chain rule", dy/dx= (dy/du)(du/dx).  We cannot prove that by simply "canceling" the two "du" terms but we can prove it by going back before the limit, doing the canceling there and taking the limit again.
Since "dy/dx" can be treated like a fraction, a notation has been developed to emphasize that.  Given y= f(x) so that dy/dx= f'(x) we define the "differential", dy, to be equal to f'(x) dx where dx, quite simply, is left as an "undefined" term but treated its as an "infinitesimal" change in x.  That allows us to do things like change "dy/dx= f(x)" to "dy= f(x)dx" and then integrate both sides.
It is simply a notational device to remind us that we can do things that otherwise would not be so obvious.
A: For your Last expression see:
Note that: $$\mathrm {d} f(x,y)=\frac{\partial f}{\partial x}.\mathrm {dx}+\frac{\partial f}{\partial y}.\mathrm {dy}$$
$$\mathrm {d} \sqrt{x^2+y^2}=\mathrm {d}(x^2+y^2)^{\frac{1}{2}}\\
=\frac{1}{2}(x^2+y^2)^{\frac{-1}{2}}\cdot\mathrm {d}(x^2+y^2)\\
=\frac{1}{2\sqrt{(x^2+y^2)}}\cdot d(x^2+y^2)$$
So, $$\pm\int  \frac{d\left(x^2+y^2\right)}{2 \sqrt{x^2+y^2}}=\pm \int \mathrm {d} \sqrt{x^2+y^2}\\
=\pm \sqrt{x^2+y^2}$$
Also, $$\mathrm {d}(x^2+y^2)=2x\mathrm {dx}+2y\mathrm {dy}$$
so,$$\frac{\mathrm {d}(x^2+y^2)}{\mathrm {dx}}=2y\frac{\mathrm {dy}}{\mathrm {dx}}+2x$$
A: A lot of this is shorthand that is justified by defining "differentials" in a certain way. A truly rigorous treatment uses methods of differential geometry, for which a good book might be Weintraub's "Differential Forms". 
The equivalence of $$\tag 1\frac{dy}{dx}=\frac{M(x,y)}{N(x,y)}$$ and $$\tag 2-M(x,y)dx+N(x,y)dy=0$$ is by no means obvious. 
$(1)$ should be clear: it's just a first order differential equation, presumably with an initial condition that is not shown. 
For $(2)$ you need some calc III:
Suppose $f:\mathbb R^{2}\to \mathbb R$ is differentiable at $\vec x=(x,y)\in R^{2}$. Let $\vec h=(\Delta x, \Delta y)$
Then $$\tag 3\lim _{\vert \vec h\vert \to 0}\left | \frac{f(\vec x+\vec h)-f(\vec x)-\nabla f(\vec x)\cdot \vec h)}{\vec h} \right |=0$$ where $\nabla f(\vec x)=\frac{\partial f}{\partial x}(\vec x)\vec i+\frac{\partial f}{\partial y}(\vec x)\vec j$.
This formula just says that if $f$ is differentiable at $\vec x$ then the tangent plane 
$$\tag 4z(\vec h)=\frac{\partial f}{\partial x}(\vec x)\Delta x+\frac{\partial f}{\partial y}(\vec x)\Delta y$$ is a good approximation to $f$ if you are near the fixed point $\vec x$.
Now, suppose $\nabla f$ is identically zero $\textit {in some neighborhood N of}$ $\vec x$. Then. $f$ is constant on $N$. 
(I'll do the proof of this fact at the end). 
So we have 
$$\tag 5f(x,y)=C$$ for some constant $C$. from which one presumably can solve for $y$ (although this requires proof of course). 
Set $\Delta x=dx;\Delta y=dy$. Then you get$$\tag 6z(\vec h)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=0$$.
Now compare $(6)$ and $(2)$ and relate this to $(5)$. When you write $(2)$ you are implicitly assuming $(6)$ and therefore $(5)$. So now we have the meaning of $(2)$: we seek a function implicitly defined by $(5)$. Luckily, the method one uses to do this will also say whether such an $f$ exists. 
Now for the proof of the claim. If $\nabla f$ vanishes on $N$, then $f$ is constant there:
A neighborhood $N$ of $\vec x$ is simply a disk, so if we pick $\vec a$ and $\vec b$ in $N$, then there is a line segment connecting $\vec a$ and $\vec b$ so we can take 
$g:[0,1]\to \mathbb R^{2}$ defined by $t\mapsto t\vec b+(1-t)\vec a$ and then consider 
$\varphi =f\circ g:[0,1]\to \mathbb R$. 
We may now apply the ordinary MVT to write 
$\varphi (1)-\varphi (0)=\varphi '(c)(1-0)$ for some $c\in (0,1)$. This is
$$\tag 1f(\vec b)-f(\vec a)=\varphi '(c)$$ But the chain rule says that 
$\varphi '(c)=f'(g(c))\cdot g'(c)=\nabla f(g(c))\cdot (\vec b-\vec a)$. Now observe that $c\in (0,1)\Rightarrow g(c)\in N$ because the entire line segment from $\vec a$ to $\vec b$ is in $N$. But by assumption $\nabla f=0$ on $N$ so in fact, $\nabla f(g(c))=0\Rightarrow \varphi '(c)=0$ and now from $(1)$ we conclude that $f(\vec b)=f(\vec a)$. As $\vec a$ and $\vec b$ were arbitrary vectors in $N$, the claim follows.
A: if this is your equation $$x^2ydy-(xdy-ydx)=0$$ i would write $$x^2y\frac{dy}{dx}-x\frac{dy}{dx}+y=0$$ or $$x^2y(x)y'(x)-xy'(x)+y(x)=0$$
