Notation Issue related to inverse function theorem

Question

So I was asked a few days ago this question: Consider $(x,y)=(f(u,v),g(u,v))$ with $\partial(x,y)/\partial(u,v)\neq0$, show that $$\left(\dfrac{\partial x}{\partial u}\right)_v\left(\dfrac{\partial u}{\partial x}\right)_y=\left(\dfrac{\partial y}{\partial v}\right)_u\left(\dfrac{\partial v}{\partial y}\right)_x$$ and the thing is, I'm actually confused about the notation. I tried with small examples assuming that the subindex means partial derivative... but it doesn't seem to work, neither I've fund useful stuff online. Any hint would very appreciated. Thanks a lot

Answer 1

There is a small trick involved. Since, $\partial(x,y)/\partial(u,v) \neq 0$, we can assume a non-zero $|\mathbf J|$ for the transformation. $$ \mathbf J = \frac{\partial(x,y)}{\partial(u,v)} = \begin{bmatrix} \partial x/\partial u & \partial x/\partial v\\ \partial y/\partial u & \partial y/\partial v \end{bmatrix} $$ It is simple to find the inverse of this matrix \begin{align} \mathbf{J}^{-1} = \frac{1}{|\mathbf J|} \begin{bmatrix} \partial y/\partial v & -\partial x/\partial v\\ -\partial y/\partial u & \partial x/\partial u \end{bmatrix} \tag{1} \end{align} But the following equation also holds by nature of inverse transformation \begin{align} \mathbf J^{-1} = \frac{\partial(u,v)}{\partial(x,y)} = \begin{bmatrix} \partial u/\partial x & \partial u/\partial y\\ \partial v/\partial x & \partial v/\partial y \end{bmatrix} \tag{2} \end{align}

You can prove your result using the equation (1)$=$(2).