In the context of curl, what does the difference between two partial derivatives tell me about rotation in a plane?

Question

Let $\mathbf{A}(x,y,z)$ be a vector field. The curl of this vector field is defined as

$$\nabla \times \mathbf{A} = \left(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}\right) \mathbf{i} + \left(\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}\right) \mathbf{j} + \left(\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\right) \mathbf{k}$$

Let's consider a simple example from Wikipedia.

$$\mathbf{A}(x,y,z) = y\mathbf{\hat{x}}-x\mathbf{\hat{y}}+0\mathbf{\hat{z}}$$

This corresponds to the following

source: https://commons.wikimedia.org/wiki/File:Uniform_curl.svg

The curl is

$$\nabla \times \mathbf{A} =0\boldsymbol{\hat{x}}+0\mathbf{\hat{y}}+ \left({\frac{\partial}{\partial x}}(-x) -{\frac{\partial}{\partial y}} y\right)\mathbf{\hat{z}}=-2\mathbf{\hat{z}}$$

But there is where I am confused. I don't get what $\left({\frac{\partial}{\partial x}}(-x) -{\frac{\partial}{\partial y}} y\right)$ has to do with the rotation and size of the vectors in the picture shown above.

My Question

How do these derivatives tell me anything about the rotation? Any why must they be opposites? Why is it not $\frac{\partial A_x}{\partial x}$? I understand of course why, when you compute the cross product, it comes out that way, but I want to understand the intuition of why that is important for rotation.