= Solution
Applying the <chain rule> to both inverse identities gives
$$
\boxed{D(f^{-1})(f(a))=[Df(a)]^{-1}.}
$$
For the unnumbered calculation, the requested point is $f(c,c)$. The Jacobian there is
$$
Df(c,c)=\begin{pmatrix}e^c&e^c\\e^c&-e^{-c}\end{pmatrix},\qquad\det Df(c,c)=-(e^{2c}+1)\ne0.
$$
To apply its inverse to $(1/2,1/2)$, solve $e^c(v_1+v_2)=1/2$, $e^cv_1-e^{-c}v_2=1/2$. Subtraction gives $v_2=0$, and hence
$$
\boxed{D_{(1/2,1/2)}f^{-1}(2e^c,2\cosh c)=\left(\frac{e^{-c}}2,0\right).}
$$
For precision, the displayed example maps bijectively onto $V=\{(s,t):s>0,t>0,st>1\}$, rather than the entire positive quadrant. Indeed writing $q=e^x$, the equations are $(s-q)(t-q)=1$ with $0<q<\min(s,t)$. This left side decreases continuously from $st$ to zero, giving exactly one $q$ if and only if $st>1$, after which $e^y=s-q$ is unique. Its Jacobian never vanishes, so the inverse is differentiable on $V$. The requested point lies in $V$, making the calculation well-defined.
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