How to prove the following? I was reading an introductory book about differential geometry, the following property seems obvious to the author but not to me :

Is this equality really obvious ? If it not, can you suggest me a strategy to prove it ? (not looking for a complete proof)
 A: Yes, this equality is obvious, but notation may sometimes be confusing. Let's say we denote the function mapping the $(\xi^i)$ to $\rho^j$ by $f^j$, that is $$ \rho^j = f^j(\xi^1, \ldots, \xi^n), \qquad j = 1, \ldots, n $$
(it's usually denoted by $\rho^j$), and the function which gives $\xi^i$ out of the $(\rho^j)$ by $g^i$, so 
$$ \xi^i = g^i(\rho^1, \ldots, \rho^n), \qquad i = 1, \ldots, n  $$
Lets write $f= (f^1, \ldots, f^n)$ and $g = (g^1, \ldots, g^n)$. We have now, converting $\xi$'s to $\rho$s and back and vice versa, that 
$$ (g^i\circ f)(\xi^1, \ldots, \xi_n) = \xi^i, \qquad (f^j\circ g)(\rho^1, \ldots, \rho^n) = \rho^j $$
That is $f \circ g = {\rm id}$, $g \circ f = {\rm id}$. Differentiating this equality gives the statement above.
A: Let $\phi$ be the coordinate transition map (i.e. such that $\rho = \phi \circ \xi$) - so the derivative of $\phi$ is $\partial \rho/\partial \xi$, and that of $\phi^{-1}$ is $\partial \xi/\partial \rho$. Apply the multivariable chain rule to the equation $\phi \circ \phi^{-1} = \phi^{-1} \circ \phi = \rm id$, remembering that the derivative of the identity map is $\partial x^i/\partial x^j = \delta^i_j$.
