relation between differential forms and standard calculus? I have been studying calculus for a long time, including multivariate calculus. 
Now, suddenly, I'm reading about the concept of a "differential form", referring to $dx$ and $dy$ and so forth. 
I've been seeing these symbols all the time in formulas like $\int f(x)dx$ and $dy=\frac {\delta f}{\delta x}dx + \frac {\delta f }{\delta y} dy$
My question is: is this term "differential form" an extension/advanced-version of the $dx$'s and $dy$'s that I've been working with, or have I been working with differential forms all this time without knowing it?
 A: Yep, you've been working with differential forms this whole time without knowing it. In fact, the things which you would call vector field were actually differential forms (although vector fields are still a thing). Scalar functions are $0$-forms, the "vector fields" that you would integrate over curves (often written as $f\,dx+g\,dy+h\,dz$) were $1$-forms, the ones you integrated over surfaces were $2$-forms, and the scalar functions you integrated over volumes can be thought of as $3$-forms (recall that the integral of a $3$-form $f\, dx\wedge dy\wedge dz$ defined on a $3$-manifold in $\mathbb R^3$ is equal to the ordinary integral of $f$ over the volume). 
Notice how the "vector fields" you integrated over curves are a different object than the "vector fields" you integrated over surfaces; the former is a $1$-form, the latter is a $2$-form. This would be more clear if we lived in a four-dimensional universe, for in that case, a $1$-form would have four components whereas a $2$-form would have ${4\choose 2}=6$ components. 
Not only that, but the gradient, curl, and divergence are all special cases of the exterior derivative. For example, let $\omega = f\, dx\wedge dy+g\,dy\wedge dz+h\, dz\wedge dx$ be a $2$-form. Then
$$d\omega = df\wedge dx\wedge dy+dg\wedge dy\wedge dz+dh\wedge dz\wedge dx$$
$$= \left(\frac{\partial f}{\partial x}+\frac{\partial g}{\partial y}+\frac{\partial h}{\partial z} \right)dx\wedge dy\wedge dz$$
If you carry out the computation, the exterior derivative of a $1$-form gives the formula for the "curl." 
You may also recall that the curl of the gradient of a scalar function is zero, as is the divergence of the curl of a "vector field". More generally, when applied to a form, the exterior derivative applied twice gives zero. 
