This might help.
You don't take the gradient of a vector field, so $\nabla A$ makes no sense. Instead, the gradient is an operator that acts on scalar functions in $\mathbb{R}^n$ and produces a vector field on $\mathbb R^n$.
Now if $f(x,y)$ is a scalar function, then $\nabla f:=\langle f_x,f_y\rangle$.
As for the Laplacian operator, it is given by $\Delta=\nabla\cdot \nabla$, i.e. the divergence of the gradient. So for example, with the $f(x,y)$ above we would have $$\Delta f:=\nabla \cdot \nabla f=\nabla \cdot \langle f_x,f_y\rangle=f_{xx}+f_{yy},$$ and similarly for higher dimensions.
The Laplacian operator is an extremely important and ubiquitous operator throughout pure and applied mathematics, certainly in areas of PDE and physics.
A simple application of why the Laplacian operator is so central is because many important problems in applied mathematics and physics can be distilled down to this:
Suppose $\mathbf F$ is a given conservative vector field (meaning
$\mathbf F=\nabla f$ for some scalar function $f$) which is also
divergence-free (meaning $\nabla\cdot\mathbf F=0$). How do we find the
$f$ (called a potential function) associated with this vector field
$\mathbf F$?
The answer is quickly revealed: $$\mathbf F=\nabla f \implies \nabla \cdot\mathbf F=\nabla\cdot \nabla f\implies 0=\Delta f.$$ The potential $f$ is a solution of Laplace's equation! (In fact, this is why some books refer to it as the potential equation.
First, writing the vectors in component form (as column vectors) has made it less obvious that you need to use the product rule here.
Let $\hat x, \hat y, \hat z$ be the usual Cartesian basis. The curl takes the form
$$\nabla \times A = (\nabla A^x) \times \hat x + A_x \nabla \times \hat x + \ldots$$
But since $\hat x$ is constant, $\nabla \times \hat x$ is zero. Hence, the formula for curl in Cartesian can be written
$$\nabla \times A = (\nabla A^x) \times \hat x + (\nabla A^y) \times \hat y + (\nabla A^z) \times \hat z$$
Once you do the cross products, you get $(\partial_y A^z - \partial_z A^y) \hat x$ and so on, as you usually would expect.
In spherical, however, the basis vectors depend on position. $\nabla \times \hat \theta$ isn't zero, for instance. That's where the terms in Wikipeida's form come from.
Second, when you converted $3x \hat x - z \hat y + 2 y \hat z$, you converted $x, y, z$ to spherical coordinates, but you didn't convert the basis. It's quite clear that you wrote
$$ 3x \hat x - z \hat y + 2 y \hat z = 3 r \sin \theta \cos \phi \hat x - r \cos \theta \hat y + 2r \sin \theta \sin \phi \hat z$$
You need to write this in terms of the spherical basis vectors in order to apply the formula for curl in spherical properly.
In summary, remember that curl in a general coordinate system is not as simple as it looks in Cartesian. You can always derive the correct formula for a given coordinate basis by using the product rule. If $\alpha$ is a scalar field and $F$ a vector field, then
$$\nabla \times (\alpha F)= (\nabla \alpha) \times F + \alpha \nabla \times F$$
make $\alpha$ the components and $F$ the basis vectors to derive the correct curl formula for your coordinate system.
Second, make sure you write all vectors and vector operators in the same basis.
Best Answer
It's best when you're working with not-true-vectors such as curl to calculate from first principles: $\frac{\partial}{\partial t}\vec{\nabla}\times\vec{v}$ has $i$th component $\epsilon_{ijk}\frac{\partial}{\partial t}\frac{\partial}{\partial x_j} v_k$, with implicit summation over repeated indices. Since the partial derivatives commute, this expression is $\epsilon_{ijk}\frac{\partial}{\partial x_j} \frac{\partial}{\partial t}v_k$, which is the $i$th component of $\vec{\nabla}\times\frac{\partial}{\partial t}\vec{v}$. In other words, one of the terms expected from a naïve Leibniz rule vanishes. Please don't interpret this as meaning $\frac{\partial}{\partial t}\vec{\nabla}=\vec{0}$, whatever that would mean.