Manipulating Partial Derivatives of Inverse Function In lectures we're told:
$$\dfrac {\partial y} {\partial x} = \dfrac 1 {\dfrac {\partial x} {\partial y}}$$
as long as the same variables are being held constant in each partial derivative.
The course is 'applied maths', i.e non-rigorous, so don't confuse me.
But anyway, if we have:
$$\xi = x - y \qquad \eta = x$$
Then $\dfrac {\partial x} {\partial \xi} = 0$ and $\dfrac {\partial \xi} {\partial x} = 1$.
The rule presumably fails because one of the partial derivatives is $0$.
But then isn't this rule useless? Since I cannot use it without checking that the partial derivative isn't in fact $0$, but then I've worked out the partial derivative manually anyway.
 A: 
The rule presumably fails because one of the partial derivatives is $0$

No, not because of that. Take 
$$\xi = x - y \qquad \eta = x+y$$
then 
$$x = \frac12(\xi+\eta) \qquad y = \frac12(\eta-\xi)$$
so $\dfrac {\partial x} {\partial \xi} = \dfrac12$ and $\dfrac {\partial \xi} {\partial x} = 1$.

The correct way to find the derivative of the inverse map is to arrange partial derivatives into a matrix, and take the inverse of that matrix. In your scenario 
$$
\begin{pmatrix}
\frac {\partial \xi} {\partial x} & \frac {\partial \xi} {\partial y}
\\
\frac {\partial \eta} {\partial x} & \frac {\partial \eta} {\partial y}
\end{pmatrix} = \begin{pmatrix}
1 & -1
\\
1 & 0
\end{pmatrix}
$$
hence
$$
\begin{pmatrix}
\frac{\partial x} {\partial \xi}  & \frac{\partial x} {\partial \eta} 
\\
\frac{\partial y}  {\partial \xi} & \frac  {\partial y}{\partial \eta}
\end{pmatrix} = \begin{pmatrix}
1 & -1
\\
1 & 0
\end{pmatrix}^{-1} = \begin{pmatrix}
0 & 1
\\
-1 & 1
\end{pmatrix}
$$
from where you can read off all the partials of the inverse map.

The rule you were told has an important clause

as long as the same variables are being held constant in each partial derivative.

which is usually the case when we take partials of the inverse map. For it to apply, you would need to compute $\dfrac {\partial x} {\partial \xi} = 0$ while holding $y$ constant (not $\eta$). That is, holding $\eta-\xi$ constant. Then the rule gives the correct result : 
$$\dfrac {\partial x} {\partial \xi}\bigg|_{\eta-\xi \text{ constant}} = 1$$ 
despite 
$$\dfrac {\partial x} {\partial \xi}\bigg|_{\eta \text{ constant}} = 0$$ 
