Why f is no longer a function of x and y ?

[Adesh]

The same difference as differentiating with respect to $y_1$ and $y_2$ in the above examples.

That solves my whole problem. Thank you so much.

 

[archaic]

So, what’s the difference between differentiating with respect to $x$ and $x’$ ?

In that response I was trying to illustrate to you the way we think of axes, but:
Define $\vec a=f(x)\vec e_1=(f(x), 0)$ and $\vec c=g(x')\vec e_1=(g(x'),0)$. You know how to do differentiation on vectors as you are reading Griffith's book, so what do you think?

That solves my whole problem. Thank you so much.

I think the main point that you need to remember is to think of the axes as the set of real numbers. The function you have of both $x$ and $x'$ who are on the same "axis" can be seen as $\vec a = f(x, x')\vec e$.

 

[mitochan]

On #4
$$\nabla\times J=[\frac{\partial J_y(x',y',z')}{\partial z}-\frac{\partial J_z(x',y',z')}{\partial y}]\hat x + ...$$
is zero.