I have explicitly known expressions for:
$f(x,y), u(x,y), v(x,y), w(x,y)$
$\frac{\partial f}{\partial x}(x,y), \frac{\partial f}{\partial y}(x,y), \frac{\partial u}{\partial x}(x,y), \frac{\partial u}{\partial y}(x,y), \frac{\partial v}{\partial x}(x,y), \frac{\partial v}{\partial y}(x,y), \frac{\partial w}{\partial x}(x,y), \frac{\partial w}{\partial y}(x,y)$
, from which I want to obtain explicit expression for:
$\frac{\partial f}{\partial u}(x,y)$ and $\frac{\partial f}{\partial v}(x,y)$
At first, I thought I should go for chain-rule, but it seems I cannot get the answer (3 unknowns from 2 equation as below)
$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}+\frac{\partial f}{\partial w}\frac{\partial w}{\partial x}$
$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial f}{\partial w}\frac{\partial w}{\partial y}$
I couldn't find anything similar to my problem (searched for multivariate derivative chain-rule, functional derivative), and couldn't come up with better keywords for the questions.
Can anyone let me know how to solve my problem, or more relevant keywords for my problem? Thank you.
P.S. For better understanding of my problem, explicit expressions are:
$u(x,y) = \sin(x)\cos(y)$
$v(x,y) = \sin(x)\sin(y)$
$w(x,y) = \cos(x)$
$f(x,y) = Y_l^m(x,y)$ is spherical harmonics
You have to pick two of $u,v,w$ to be free variables (noting that $u^2+v^2+w^2=1$, whence all three of them cannot be treated as free variables). I am guessing that $u$ and $v$ are free. However, the assignment $(x,y)\mapsto (u,v)$ is an almost everywhere $4$-to-$1$ map from $(\mathbb{R}/2\pi\mathbb{Z})\times(\mathbb{R}/2\pi\mathbb{Z})$ to $[-1,+1]\times[-1,+1]$. In other words, $$u(x,y)=u(-x,y+\pi)=u(x+\pi,y+\pi)=u(-x+\pi,y)$$ and $$v(x,y)=v(-x,y+\pi)=v(x+\pi,y+\pi)=v(-x+\pi,y)\,,$$ where the operations $z\mapsto -z$ and $z\mapsto z+\pi$ are considered modulo $2\pi$. Consequently, it is quite problematic to go from $(u,v)$ back to $(x,y)$. However, if $x$ and $y$ are both known, then you can see that $$x=s\,\text{arcsin}\big(\sqrt{u^2+v^2}\big)+t\text{ and }y=\text{arctan}\left(\frac{v}{u}\right)+\tau\,,$$ where $s\in\{-1,+1\}$ and $t,\tau\in\{0,\pi\}$ (indeed, $s$ is the sign of $\sin(x)$). The values of $s$, $t$, and $\tau$ are fixed in a small neighborhood of each noncritical point. That is, $$\frac{\partial f}{\partial u}(x,y)=\left(\frac{\partial f}{\partial x}(x,y)\right)\,\left(\frac{\partial x}{\partial u}\right)+\left(\frac{\partial f}{\partial y}(x,y)\right)\,\left(\frac{\partial y}{\partial u}\right)$$ and $$\frac{\partial f}{\partial v}(x,y)=\left(\frac{\partial f}{\partial x}(x,y)\right)\,\left(\frac{\partial x}{\partial v}\right)+\left(\frac{\partial f}{\partial y}(x,y)\right)\,\left(\frac{\partial y}{\partial v}\right)$$ can be determined from $$\frac{\partial x}{\partial u}=\frac{su}{\sqrt{(1-u^2-v^2)(u^2+v^2)}}=\frac{\cos(y)}{\big|\cos(x)\big|}\,,$$ $$\frac{\partial x}{\partial v}=\frac{sv}{\sqrt{(1-u^2-v^2)(u^2+v^2)}}=\frac{\sin(y)}{\big|\cos(x)\big|}\,,$$ $$\frac{\partial y}{\partial u}=-\frac{v}{u^2+v^2}=-\frac{\sin(y)}{\sin(x)}$$ and $$\frac{\partial y}{\partial v}=+\frac{u}{u^2+v^2}=+\frac{\cos(y)}{\sin(x)}\,.$$