I'm focused on the expression
$$ \left(f(x,y) \frac{d}{dx} \right)^n$$ as an operator, and I'd like to calculate its partial derivatives when applied to a sufficiently differentiable function.
Can I assert that
$$ \frac{d}{dy} \left(f(x,y) \frac{d}{dx} \right)^n[g(x)] = \left( n\left(f(x,y) \frac{d}{dx} \right)^{n-1} \cdot f_y \frac{d}{dx} \right)[g(x)]?$$
Or is this generally the incorrect formula? If that's incorrect, the alternative I suspect is
$$\left( n\left(f(x,y) \frac{d}{dx} \right)^{n-1} \cdot \left(f_y \frac{d}{dx} + f \frac{d}{da} \frac{d}{dx} \right) \right)[g(x)]?$$
However, another poster brought up a fair point, which is to look at the case of $n = 1$ first. Because $\frac{d}{dx}$ is not a "function" though, I'm unsure whether my interpretation is correct for $n=1$ too.
What is $$\frac{d}{dx} \left( f(x,y) \frac{d}{dx} \right) \ \text{and} \ \frac{d}{dy} \left(f(x,y) \frac{d}{dx} \right)?$$
No. It's not so easy as applying the power rule.
$\left[f(x,y)\dfrac{\partial~~}{\partial x}\right]^n$ is not the $n$-th exponential of a product of functions, but rather it is the $n$-th composition of product of a function and a linear operator.
Thus the product rule for derivation shall be involved so many times.
Examine a simple case:
$$\small\begin{align}\left[f(x,y)\tfrac{\partial~~}{\partial x}\right]^2&= \left[f(x,y) \tfrac{\partial~~}{\partial x}\right]\!\left[f(x,y)\tfrac{\partial~~}{\partial x}\right] \\ &=\left[f(x,y)\tfrac{\partial f(x,y)}{\partial x}\tfrac{\partial ~~}{\partial x}+f(x,y)^2\tfrac{\partial^2~~~}{\partial x~^2}\right]\\[2ex]\tfrac{\partial~~}{\partial y}\left[f(x,y)\tfrac{\partial~~}{\partial x}\right]^2&=\left[\tfrac{\partial f(x,y)}{\partial y}\tfrac{\partial f(x,y)}{\partial x}\tfrac{\partial~~}{\partial x}+f(x,y)\tfrac{\partial^2 f(x,y)}{\partial y~\partial x}\tfrac{\partial~~}{\partial x}+f(x,y)\tfrac{\partial f(x,y)}{\partial x}\tfrac{\partial^2~~}{\partial y~\partial x}+2f(x,y)\tfrac{\partial f(x,y)}{\partial y}\tfrac{\partial ~~}{\partial x}+f(x,y)^2\tfrac{\partial^2~~}{\partial y~\partial x}\right]\\[2ex]\tfrac{\partial~~}{\partial y}\left[f(x,y)\tfrac{\partial~~}{\partial x}\right]^2 g(x)&=\tfrac{\partial f(x,y)}{\partial y}\tfrac{\partial f(x,y)}{\partial x}\tfrac{\mathrm d g(x)}{\mathrm d x}+f(x,y)\tfrac{\partial^2 f(x,y)}{\partial y~\partial x}\tfrac{\mathrm d g(x)}{\mathrm d x}+2f(x,y)\tfrac{\partial f(x,y)}{\partial y}\tfrac{\mathrm d g(x)}{\mathrm d x}\end{align}$$