= Solution
The <inverse function theorem> states: if $f$ is continuously differentiable on an open subset of $\mathbb R^n$ and $Df(a)$ is invertible, then some <open neighbourhood> of $a$ is mapped bijectively onto an <open neighbourhood> of $f(a)$, with continuously differentiable inverse. Its derivative at $f(a)$ is $(Df(a))^{-1}$.
The present map is smooth, and at $a=(\pi/2,\pi)$,
$$
Df(a)=\begin{pmatrix}0&-e^{\pi/2}\\-e^{-\pi}&0\end{pmatrix},\qquad
\det Df(a)=-e^{-\pi/2}\ne0.
$$
Also $f(a)=(0,0)$, so the theorem applies locally and gives
$$
\boxed{D(f^{-1})(0,0)=\begin{pmatrix}0&-e^{\pi}\\-e^{-\pi/2}&0\end{pmatrix}}.
$$
The inverse is defined near the image point $(0,0)$ and takes values near $a$; no assertion of a global inverse is required.
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