Find the jacobian I'm been struggling with the problem for a quite some time now.
I need to find the jacobian for the following : 
$$u=x-y$$
$$v=xy$$
What I did :
$$x=y+u\\x=\frac{v}{y}\\y=x-u\\y=\frac{v}{x}$$
\begin{vmatrix}
dx/du & dx/dv \\ 
dy/du & dy/dv 
\end{vmatrix} 
\begin{vmatrix}
1 & \frac{1}{y} \\ 
-1 & \frac{1}{x} 
\end{vmatrix} 
$$\frac{1}{x}+\frac{1}{y}$$
But for some reason the answer say I should get : $$\frac{1}{x+y}$$
The method they used in the book is to calculate $J^{-1}$ which is $x+y$ and use the law $J(x,y)*J(u,v)=1$ in order to get $J(u,v)=\frac{1}{x+y}$
I don't know, why I get different answer.
Any idea?
Any help will be appreciated, Thanks in advance!
 A: The simple mnemonic is this:
When ever you are computing the Jacobian, elements of the same coordinate system must go on the same side of the equality sign
What you did wrong: writing $x = y + u$ etc. 

When computing $\partial x / \partial u$ in computing the Jacobian, you are asked to compute "the partial derivative of $x$ relative to varying $u$ while keeping $v$ fixed, when $x$ is seen as a function of $u$ and $v$." 
So the expression 
$$ x = y + u$$
really should mean 
$$ x(u,v) = y(u,v) + u $$
since you need to express the function $x$ as a function of $u$ and $v$. So when you compute
$ \partial x / \partial u $ what you should end up with from the equation above is
$$ \frac{\partial x}{\partial u} = \frac{\partial y}{\partial u} + 1 $$
(you arrived at 1, which is the result of "the partial derivative of $x$ with respect to varying $u$ if $y$ is kept fixed", since $y$ and $v$ are not the same, what you computed is not the correct thing). 

In these kinds of situations a helpful book keeping trick is the one that physicists use often in thermodynamics:
instead of writing $\partial x / \partial u$, you can write $\frac{\partial x}{\partial u} |_{v}$ to remind yourself that the partial derivative with respect to $u$ is taken with $v$ held fixed, and nothing else. So when computing, when $\partial_u$ acts on anything that is not $u$, and $v$, you must apply the chain rule. 

Edit: after saying what you shouldn't do, let me say what you should do. 
Starting with 
$$ x + y = u$$
and
$$ xy = v$$
Taking the partials relative to $u$ (while holding $v$ fixed!!!) is 
$$ \frac{\partial x}{\partial u} + \frac{\partial y}{\partial u} = 1 $$
$$ \frac{\partial x}{\partial u} y + x \frac{\partial y}{\partial u} = 0$$
and taking the partials relative to $v$ you get
$$ \frac{\partial x}{\partial v} + \frac{\partial y}{\partial v} = 0 $$
$$ \frac{\partial x}{\partial v} y + x \frac{\partial y}{\partial v} = 1 $$
which, treating $x$ and $y$ as "knowns", is a system of four equations for four unknowns. Solving this system you will get the partials as functions of $x,y$ and from that you can compute the Jacobian. As you can see now that the method of computing $J^{-1}$ is computationally simpler. 
A: Solution (Alternative Way):
We were given the transformation
$$u=x-y$$
$$v=xy$$
Notice that rewriting the equations in terms of $x$ and $y$ will make things harder (since you cannot solve for $x$ and $y$ explicitly.) However, we invoke the idea that
$$J(u,v) J(x,y)=1$$
Thus instead of finding $J(u,v)$, we try to compute $J(x,y)$ since we already have explicit expressions for $u$ and $v$.
$$J(x,y)=\left | \frac{\partial(u,v)}{\partial(x,y)} \right |=\begin{vmatrix}
u_x &u_y \\ 
v_x &v_y 
\end{vmatrix}=\begin{vmatrix}
1 &-1 \\ 
y &x 
\end{vmatrix}=x-(-y)=x+y$$
Therefore
$$J(u,v)=\frac{1}{J(x,y)}=\frac{1}{x+y}$$
