Deriving the Vector Potential

We have motivated the derivation of the vector potential in the following way:

However, I cannot understand where the ##-## sign in the second equality came from. I thought that it was because the gradient was with respect to the ##y##-variable, and then using the product rule one could explain the transition to the last expression, but in that case ##\nabla_{\vec y}\times\vec j(\vec y)## would have to be zero, which I am not really sure is necessarily true; and in that case I would again not understand how a ##\nabla_{\vec y}## would become a ##\nabla_{\vec x}##, since at ##\nabla\times \vec A(\vec x)## I assume ##\nabla## must be acting on the ##\vec x##
That's why I don't see how the left and right hand sides of the third, fourth, and possibly the fifts ##=## signs are equal to each other, can someone please help me?

You know that
##\nabla_x \dfrac{1}{|x-y|}=-\dfrac{x-y}{|x-y|^3}=-\nabla_y \dfrac{1}{|x-y|}##
It follows that
##\dfrac{x-y}{|x-y|^3}=+\nabla_y \dfrac{1}{|x-y|}=-\nabla_x \dfrac{1}{|x-y|}.##

So the second equality is $$=-\frac{1}{c}\int \int \int j(y)\times\nabla_x \dfrac{1}{|x-y|}d^3y$$Does this help?

but in that case how do we take ##\nabla_{\vec x}## out of the integral? It wasn't cross multiplied with ##\frac 1 {|\vec x-\vec y|}##, but now it is?

The vector identity says
##\vec{\nabla}\times(\psi~\vec A)= \psi \vec{\nabla}\times\vec A+\vec{\nabla}\psi\times \vec A. ##
Here you identify
##\vec{\nabla}\rightarrow \vec{\nabla}_x##
##\psi \rightarrow \dfrac{1}{|x-y|}##
##\vec A \rightarrow \vec j (y)##

What do you get when you put these in the identity?