What does the partial derivative notation $\frac{\partial (f,g)}{\partial (x,y)}$ mean? I am currently reading an old math book which contains the following unexplained notation:
Let $f(x,y)$ and $g(x,y)$ be functions $\mathbb{R}^2\rightarrow \mathbb{R}$. The notation
$$\frac{\partial(f,g)}{\partial(x,y)}$$
apparently refers to a function of the form $\mathbb{R}^2\rightarrow \mathbb{R}$. I am not sure, but I suspect it may be defined as
$$\frac{\partial(f,g)}{\partial(x,y)} \equiv \frac{\partial f}{\partial x}\frac{\partial g}{\partial y}-\frac{\partial f}{\partial y}\frac{\partial g}{\partial x}.$$
Is this notation/definition common in any particular field?
And especially if so, could there be an obvious interpretation of the following notation, which this book also uses without explanation?
$$\frac{\partial[f,g]}{\partial(x,y)}$$
Thanks for your help.

Edit: I believe @Fred is correct that the parentheses are used to denote the Jacobian.
Here is the notation, as used in a simplified excerpt of Calculating Curves by Ron Doerfler and others:


$$
\left\{ \begin{array}{c} 0 = \frac{\partial u}{\partial y} f_1(x) +
\frac{\partial v}{\partial y}\\ 0 = \frac{\partial u}{\partial x}
 f_2(y) + \frac{\partial v}{\partial x}\\ \end{array}\right.$$
Let us assume that $\frac{\partial(u,v)}{\partial(x,y)}=0$. Then the
  above equations yield that 
$$\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}[f_1(x) -
 f_2(y)] = 0.$$
Thus we can posit that 
$$\frac{\partial(u,v)}{\partial(x,y)} = \frac{\partial u}{\partial
x}\frac{\partial v}{\partial y}-\frac{\partial v}{\partial
 x}\frac{\partial u}{\partial y} = e^\theta.$$

I am still unsure about the meaning of the square bracket notation. The  square brackets are a little more difficult to place in context, but here is an attempt:

$$g_3(z) = u f_3(z) + v$$
We clearly have $\frac{\partial[g_3(z), z]}{\partial(x,y)} = 0$. By substituting the above equation and observing that $\frac{\partial[f_3(z), z]}{\partial(x,y)} = 0$, we obtain
$$f_3(z)\frac{\partial(u,z)}{\partial(x,y)} + \frac{\partial(v,z)}{\partial(x,y)} = 0.$$

Possibly the square brackets are simply an alias for round brackets which are used to avoid potentially visually-noisy nested round brackets(?).
 A: In vector calculus, we said, for the generalization of the chain rule, that
$$\frac{\partial \mathbf f}{\partial \mathbf x}=\frac{\partial \mathbf f}{\partial \mathbf u}\frac{\partial \mathbf u}{\partial \mathbf x}$$
Where $\mathbf x, \mathbf f, \mathbf u$ are vectors (of functions).
Thus,
$$\frac{\partial (f, g)}{\partial (x, y)}=
\begin{bmatrix}
\frac{\partial f}{\partial x}\\
\frac{\partial g}{\partial x}
\end{bmatrix}
\begin{bmatrix}
\frac{\partial x}{\partial x}&
\frac{\partial x}{\partial y}
\end{bmatrix}
=
\begin{bmatrix}
\frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}\\
\frac{\partial g}{\partial x}
&\frac{\partial g}{\partial y}
\end{bmatrix}$$
Well, it is true that sometimes books use this notation for the determinant of this matrix (However I didn't). Then we get,
$$
\begin{vmatrix}
\frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}\\
\frac{\partial g}{\partial x}
&\frac{\partial g}{\partial y}
\end{vmatrix}
=
\frac{\partial f}{\partial x}\frac{\partial g}{\partial y}-
\frac{\partial f}{\partial y}
\frac{\partial g}{\partial x}
$$
Which is the Jacobian of $h(x,y)=(f(x,y), g(x,y))$.
A: Let $h: \mathbb R^2 \to  \mathbb R^2 $ be defined by $h(x,y) =(f(x,y),g(x,y))$.
Then $\frac{\partial(f,g)}{\partial(x,y)} $ is the Jacobian of $h$.
A: $$\frac{\partial(f,g)}{\partial(x,y)}
= \left(\begin{matrix}
\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\
\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}
\end{matrix}\right)$$
However, I think that sometimes the notations is used for the determinant if this matrix.
