Show $(\frac{\partial f}{\partial x})^2-(\frac{\partial f}{\partial y})^2=(\frac{\partial f}{\partial r})^2-1/{r^2}(\frac{\partial f}{\partial t})^2$ I have been asked to prove the following relation: $$(\frac{\partial f}{\partial x})^2-(\frac{\partial f}{\partial y})^2=(\frac{\partial f}{\partial r})^2-\frac 1{r^2}(\frac{\partial f}{\partial t})^2$$
Given that $x=r\cosh (t)$ and $y=r\sinh(t)$
From this it can be deduced that: 
$f=f(x,y)$,
$f=f(r,t)$,
$x=x(r,t)$,
$y=y(r,t)$,
$r=r(x,y)$ and
$t=t(x,y)$ 
Each of the above ($df,dx,dy,dt...$) will have a total derivative associated with them. Thus it if possible to use the chain rule to obtain expressions for the partial derivatives given in the question.
$$(\frac{\partial f}{\partial r})=(\frac{\partial f}{\partial x})(\frac{\partial x}{\partial r})+(\frac{\partial f}{\partial y})(\frac{\partial y}{\partial r})$$
$$(\frac{\partial f}{\partial t})=(\frac{\partial f}{\partial x})(\frac{\partial x}{\partial t})+(\frac{\partial f}{\partial y})(\frac{\partial y}{\partial t})$$
$$(\frac{\partial f}{\partial x})=(\frac{\partial f}{\partial r})(\frac{\partial r}{\partial x})+(\frac{\partial f}{\partial t})(\frac{\partial t}{\partial x})$$
$$(\frac{\partial f}{\partial y})=(\frac{\partial f}{\partial r})(\frac{\partial r}{\partial y})+(\frac{\partial f}{\partial t})(\frac{\partial t}{\partial y})$$
Now here's my issue. The problem should be possible to solve by finding equations for the squares of $(\frac{\partial f}{\partial r})$ and $(\frac{\partial f}{\partial t})$ and finding thier difference. Or by doing the same with $(\frac{\partial f}{\partial x})$ and $(\frac{\partial f}{\partial y})$. I found it very simple to do the former but I could not find the relation using the expressions for the derivatives with respect to $x$ and $y$. 
I used the following derivatives in my failed attempt using the expressions for the derivatives with respect to $x$ and $y$:
$(\frac{\partial r}{\partial x})=\cosh(t)$, $(\frac{\partial r}{\partial y})=\sinh(t)$, $(\frac{\partial t}{\partial x})=-\frac{\sinh(t)}r$ and$(\frac{\partial t}{\partial y})=\frac{\cosh(t)}r$
How is it possible?
 A: to go the other way round,
$$
f_x=f_r r_x + f_t t_x \\
f_y=f_r r_y + f_t t_y
$$
since 
$$
r^2 = x^2-y^2
$$
we have
$$
r_x= \frac{x}{r} \\
r_y= -\frac{y}{r}
$$
also
$$
t = \tanh^{-1}\frac{y}{x}
$$
so
$$
t_x = \frac{y}{r^2} \\
t_y = - \frac{x}{r^2}
$$
this gives:
$$
(f_x)^2 - (f_y)^2 = (\frac{x}{r}f_r +\frac{y}{r^2}f_t)^2 -(-\frac{y}{r}f_r -\frac{x}{r^2}f_t)^2 \\
= (f_r)^2 - \frac1{r^2} (f_t)^2
$$
using the same identity as before, and again with cancellation of the mixed terms, which this time are $\frac{2xyf_rf_t}{r^3}$
it is worth noting that the symmetry shown by the relations:
$$
\begin{pmatrix} f_r \\f_t \end{pmatrix}=\begin{pmatrix}x_r & y_r \\ x_t & y_t
\end{pmatrix} \begin{pmatrix} f_x \\ f_y \end{pmatrix}
$$
$$
\begin{pmatrix} f_x \\f_y \end{pmatrix}=\begin{pmatrix}r_x & t_x \\ r_y & t_y
\end{pmatrix} \begin{pmatrix} f_r \\ f_t \end{pmatrix}
$$
is expressed in terms of the Jacobian determinants:
$$
\frac{\partial(x,y)}{\partial(r,t)} \frac{\partial(r,t)}{\partial(x,y)} = 1
$$
A: use the abbreviations $c=\cosh t, s = \sinh t$ and denote differentiation by a subscript.
then, as you have remarked:
$$
f_r = f_x x_r + f_y y_r = f_x c + f_y s \\
f_t = f_x x_t + f_y y_t = r(f_x s + f_y c)
$$
this gives:
$$
(f_r)^2 - \frac1{r^2}(f_t)^2 = (f_x c + f_y s)^2 - (f_x s + f_y c)^2
$$
which simplifies to $(f_x)^2 - (f_y)^2$ on account of the identity $c^2-s^2 =1$ and the cancellation of the mixed terms $2f_xf_y c s$
