If differentials aren't supposed to be treated like fractions, how come we do just that in Differential Equations? For years I've heard that differentials aren't fractions and multiplying $\frac {d y}{d x} $ with $d x$ to "cancel out the denominator" isn't the right view, yet I've never been shown this "right view" and in my ODE class we always do exactly what we were told was wrong.
What exactly are the properties and proper usages of differentials? How come we're told they don't technically "cancel out" in fractions, yet nearly everything we've been exposed to counters that?
 A: Consider the following example
\begin{align}
\frac{dy}{dx} = xy
\end{align}
which is a separable differential equation. In particular, it is taught in most ODE courses that you could rewrite the equation as
\begin{align}
\frac{1}{y}dy = xdx
\end{align}
then integrate both sides accordingly and everything will workout. However, a more rigorous way is to view as follow
\begin{align}
\int \frac{1}{y}\frac{dy}{dx}\ dx = \int x\ dx
\end{align}
then by the change of variable formula we get that
\begin{align}
\int\frac{1}{y} \frac{dy}{dx}\ dx = \int \frac{1}{y}\ dy.
\end{align}
So by treating the ordinary derivative as a fraction, you simply save the step for justifying the change of variable. 
A: Given Co-Ordinate projections in ${\mathbb{R}^2}$
\begin{gathered}
  x:{\mathbb{R}^2} \to \mathbb{R},x\left( p \right) = u \hfill \\
  y:{\mathbb{R}^2} \to \mathbb{R},y\left( p \right) = v \hfill \\ 
\end{gathered} 
point $p \in {\mathbb{R}^2}$ can now be written like
$$p = \left( {u,v} \right)$$.
Differentials from these Co-Ordinate functions at point $p$
\begin{gathered}
  d{x_p} = {T_p}{\mathbb{R}^2} \to \mathbb{R} \hfill \\
  d{y_p} = {T_p}{\mathbb{R}^2} \to \mathbb{R} \hfill \\ 
\end{gathered} 
are Co-Vectors, that means $d{x_p},d{y_p} \in T_p^*{\mathbb{R}^2}$.
For every Tangent-Vector like $\xi  \in {T_p}{\mathbb{R}^2}$, we have
$$\xi  = a\frac{\partial }{{\partial x}} + b\frac{\partial }{{\partial y}}$$
$a,b \in \mathbb{R}$. Now
$$d{x_p}\left( \xi  \right) = d{x_p}\left( {a\frac{\partial }{{\partial x}} + b\frac{\partial }{{\partial y}}} \right) = d{x_p}\left( {a\frac{\partial }{{\partial x}}} \right) + d{x_p}\left( {b\frac{\partial }{{\partial y}}} \right) = a\frac{{\partial x}}{{\partial x}} + b\frac{{\partial x}}{{\partial y}} = a$$
$$d{y_p}\left( \xi  \right) = d{y_p}\left( {a\frac{\partial }{{\partial x}} + b\frac{\partial }{{\partial y}}} \right) = d{y_p}\left( {a\frac{\partial }{{\partial x}}} \right) + d{y_p}\left( {b\frac{\partial }{{\partial y}}} \right) = a\frac{{\partial y}}{{\partial x}} + b\frac{{\partial y}}{{\partial y}} = b$$.
That is: $d{x_p},d{y_p} \in T_p^*{\mathbb{R}^2}$ are linear real-valued functions. And from this point of view we are calculating with real numbers.
So we can divide them like $\frac{{d{y_p}}}{{d{x_p}}}$ and there is nothing to
worry about. If $f:{\mathbb{R}^2} \to \mathbb{R},y = f\left( x \right)$
for a differentiable function, we have
$$\frac{{d{y_p}}}{{d{x_p}}} = f'\left( p \right)$$
Nothing, but a real number.
