The transformation of $\frac{\partial y}{\partial x}=\frac{Q}{P}$ to $\frac{dy}{Q}=\frac{dx}{P}$ I'm studying a method used to solve PDE's and at one point in the argument we have the following: $$\frac{\partial y}{\partial x}=\frac{Q}{P}$$ which is then later re-written as $$\frac{dy}{Q}=\frac{dx}{P}$$ I sort of see how re-arranging fractions would make this true but I am unconvinced and something seems odd about the partials becoming "not partial". 
Can someone try to explain why this should be true?
 A: Actually, it is more a way to easily remember the method (and I guess you are talking about separable differential equation here).
We have
$$\begin{align}
\frac{\partial y}{\partial x}&=\frac{Q(y(x,z))}{P(x)}\\
\frac{\frac{\partial y}{\partial x}}{Q(y(x,z))} &=\frac{1}{P(x)}\\
\int \frac{\frac{\partial y}{\partial x}}{Q(y(x,z))}\text{d}x &= \int \frac{1}{P(x)}\text{d}x\\
\int \frac{1}{Q(u)}\text{d}u &=\int \frac{1}{P(x)}\text{d}x\\
\end{align}$$
where $u=y(x,z)$ (this substitution is the key of the method because we get rid of the $\partial y/\partial x$ and it motivates the shorthand $(\dots)\text{d}y=[\dots]\text{d}x$)
For example, if we take $y(x,z)$ with $Q(y(x,z))=-(y(x,z))^{2}$ and $P(x)=e^{x}$, it gives
$$\int \frac{1}{-u^{2}}\text{d}u =\int \frac{1}{e^{x}}\text{d}x$$
Hence
$$\begin{align}
\frac{1}{u} &=-e^{-x}+K(z)\\
u &=\frac{1}{K(z)-e^{-x}}\\
y(x,z) &=\frac{1}{K(z)-e^{-x}}
\end{align}$$
where $K(z)$ is a real function of $z\in\mathbb{R}^{n}$.
Separating the "pseudo-ratio" $\frac{\partial y}{\partial x}$ is a way to remember how to use the method, but I don't think it is a formal way to write it.
