I'm trying to understand the partial derivative of a two argument function. I understand the concept of the first PD, where one variable is held constant $f_{y}(x) = f(x, y)$, while the derivative of the single variable is taken $f'_a(x)$ where $a$ is the value of the constant. For the second PD, the same processes of creating an index family of functions is followed.
My confusion is in regards to the second PD and I have some questions that follow.
The PD at a point is also defined to be:
$$f_{a_{1},...,a_{i-1},a_{i+1},...,a_n}(x_i) = f(a_1,...,a_{i-1},x_i, a_{i+1},...,a_n)$$
\begin{align*} \frac {\partial f} {\partial x_i} (a_1,...,a_n) &= \lim_{h \to 0} \frac {f(a_1,...,a_i + h, ..., a_n) - f(a_1,..., a_i,, ..., a_n)} h\\ &= \frac {d f_{a_1,...,a_{i-1},a_{i+1},...,a_n}}{d x_i}(a_i) \end{align*}
The point form equations provide an insightful way of looking at the PD, however I'm confused when a second PD is taken $\frac {\partial^2} {\partial x \partial y} f$.
\begin{align*} \frac {\partial f} {\partial x_i} (a_1,...,a_n) &= \lim_{h \to 0} \frac {\frac \partial {\partial x_j} f(a_1,...,a_i + h, ..., a_n) - \frac \partial {\partial x_j} f(a_1,..., a_i,, ..., a_n)} h\\ &= \frac {d \frac \partial {\partial x_j} f_{a1,...,a_{i-1},a_{i+1},...,a_n}}{d x_i}(a_i) \end{align*}
Does this notation appear correct? Intuitively what does this mean? Is $\frac {\partial^2} {\partial x \partial y} f$ the change of the PD $\frac \partial {\partial y} f$ with respect to $x$? Why would we want to know this when it seems the interesting change of $f$ is $\frac \partial {\partial x} f$ and $\frac \partial {\partial y} f$?
Finally, the following example is presented. It appears $\frac{\partial}{\partial y} \frac{\partial f}{\partial x}$ and $\frac{\partial}{\partial x} \frac{\partial f}{\partial y}$ have the same value. Is this expected for all values where the order of the partial in Leibniz notation is taken or just a coincidence? I ask because I'm curious if commutative laws of algebra apply to this. Thanks!
$$f(x,y) = \sin(x) y^2$$
$$(f_y)_x = \frac{\partial}{\partial y} \frac{\partial f}{\partial x} = \frac{\partial^2 f}{\partial y \partial x} = \frac\partial{\partial y} \left(\frac \partial {\partial x} (\sin(x) y^2)\right) = \frac \partial {\partial y} (\cos(x)y^2) = 2 \cos(x)y$$
$$(f_x)_y = \frac{\partial}{\partial x} \frac{\partial f}{\partial y} = \frac{\partial^2 f}{\partial x \partial y} = \frac\partial{\partial x} \left(\frac \partial {\partial y} (\sin(x) y^2)\right) = \frac \partial {\partial x} (2\sin(x)y) = 2 \cos(x)y$$
