Let $x=f(u,v), y=g(u,v)$ such that the conditions for Inverse function theorem exists.
Prove $ \frac{\partial x}{\partial u}\cdot \frac{\partial u}{\partial x}= \frac{\partial y}{\partial v}\cdot \frac{\partial v}{\partial y} , \frac{\partial x}{\partial v}\cdot \frac{\partial v}{\partial x}= \frac{\partial y}{\partial u}\cdot \frac{\partial u}{\partial y}$
Using Inverse function theorem , is it correct that $u=f^{-1}(x,y), v=g^{-1}(x,y)$ ?
Then , I have to calculate $\frac{\partial x}{\partial u} , \frac{\partial u}{\partial x} , \frac{\partial y}{\partial v} , \frac{\partial v}{\partial y} ,\frac{\partial x}{\partial v} , \frac{\partial v}{\partial x} , \frac{\partial y}{\partial u} , \frac{\partial u}{\partial y} $ ?
I am not enitrely sure since how to approach the problem , appreciate any help.
$$(x,y)=F(u,v):=\left(f(u,v),g(u,v)\right).$$ You don't have "$u=f^{-1}(x,y),v=g^{-1}(x,y)$" but $$(u,v)=F^{-1}(x,y)=:\left(\varphi(x,y),\psi(x,y)\right).$$ You don't have to "calculate $\frac{\partial x}{\partial u} , \frac{\partial u}{\partial x} , \frac{\partial y}{\partial v} , \frac{\partial v}{\partial y} ,\frac{\partial x}{\partial v} , \frac{\partial v}{\partial x} , \frac{\partial y}{\partial u} , \frac{\partial u}{\partial y}$" (and you cannot, since the two functions $f,g$ are not known). What you must prove is that $\frac{\partial x}{\partial u}\frac{\partial u}{\partial x}= \frac{\partial y}{\partial v}\frac{\partial v}{\partial y}$ and $\frac{\partial x}{\partial v}\frac{\partial v}{\partial x}= \frac{\partial y}{\partial u}\frac{\partial u}{\partial y},$ i.e. $$\frac{\partial f}{\partial u}\left(\varphi(x,y),\psi(x,y)\right)\frac{\partial \varphi}{\partial x}(x,y)=\frac{\partial g}{\partial v}\left(\varphi(x,y),\psi(x,y)\right)\frac{\partial\psi}{\partial y}(x,y)$$ and $$\frac{\partial f}{\partial v}\left(\varphi(x,y),\psi(x,y)\right)\frac{\partial \psi}{\partial x}(x,y)=\frac{\partial g}{\partial u}\left(\varphi(x,y),\psi(x,y)\right)\frac{\partial\varphi}{\partial y}(x,y).$$ Hint to prove this: differentiate $F\circ F^{-1}={\rm id}$ and $F^{-1}\circ F={\rm id}.$