Before getting started, let us go back on some definitions:
Definition. Let $U$ be an open subset of $\mathbb{R}^n$ and let $f\colon U\rightarrow V\subseteq\mathbb{R}^n$ be a $C^1$-mapping, then $f$ is a $C^1$-diffeomorphism or globally invertible if and only if there exists $g\colon V\rightarrow U$ a $C^1$-mapping such that:
$$f\circ g=\textrm{id}_{V}\textrm{ and }g\circ f=\textrm{id}_U.$$
In other words, $f$ is a bijection whose inverse is smooth. This is not to be confused with:
Definition. The mapping $f$ is said to be a local diffeomorphism or locally invertible if and only if when retricted to an open subset of $U$, $f$ is a diffeomorphism onto its image.
Please notice that being globally invertible does not mean being locally invertible around any point.
As a reminder, let us state the inverse function theorem once again:
Theorem. Let $U$ be an open subset of $\mathbb{R}^n$, let $x$ be a point of $U$ and let $f\colon U\rightarrow\mathbb{R}^n$ be a $C^1$-mapping. Assume that $\mathrm{d}_xf\colon\mathbb{R}^n\rightarrow\mathbb{R}^n$ is an invertible linear map, then there exists $V$ an open neighbourhood of $x$ such that $f\colon V\rightarrow f(V)$ is a $C^1$-diffeomorphism.
In our case, for all $(x,y)\in\mathbb{R}^2$, the matrix of $\mathrm{d}_{(x,y)}f$ in the canonical basis of $\mathbb{R}^2$ is:
$$\begin{pmatrix}e^x\cos(y)&-e^x\sin(y)\\e^x\sin(y)&e^x\cos(y)\end{pmatrix}.$$
Its determinant is $e^{2x}$ which is nonzero for all $(x,y)$. Theorefore, according to the theorem, for all $(x,y)\in\mathbb{R}^2$, $f$ is a $C^1$-diffeomorphism in a neighborhood of $(x,y)$.
Assume by contradiction that there exists $g\colon\mathbb{R}^2\rightarrow\mathbb{R}^2$ a $C^1$-mapping such that:
$$g\circ f=\textrm{id}_{\mathbb{R}^2}.$$
Then, let $(X,Y)\in(\mathbb{R}^2)^2$ such that $f(X)=f(Y)$, then $g(f(X))=g(f(Y))$ i.e. $X=Y$ and $f$ is injective. However, $f$ is non injective since $f(0,0)=(1,0)=f(0,2\pi)$ and $(0,0)\neq (0,2\pi)$, a contradiction. Whence, $f$ is not a $C^1$-diffeomorphism.
Maybe it will give more insight to notice that if one identifies $\mathbb{R}^2$ with $\mathbb{C}$ through $(x,y)\mapsto x+iy$, then the considered mapping is the complex exponential, $z\mapsto e^z$ which is invertible on any horizontal strip of length at most $2\pi$.
It looks like you're on the right lines, though I'm not entirely sure if you typeset out the Jacobian out correctly with what you had in mind. It should be
$$ d_{(x,y)}f =
\begin{bmatrix}
e^{x}\cos y & e^{x}\sin y\\
-e^{x}\sin y & e^{x}\cos y
\end{bmatrix}
$$
which still has $\det d_{(x,y)}f = e^{x} \neq 0, $ for all $(x,y) \in \mathbb{R}^{2}$ like you said. I also agree with your reasoning between why $f$ is not globally invertible since it provides a counterexample, and also this looks like a question which acts as a hint for the next part.
$f$ is not globally invertible because of the periodic nature, so we can just restrict the domain from $\mathbb{R} \times \mathbb{R}$ to $A := \mathbb{R} \times (0, 2\pi )$, which remains surjective once one removes the origin from the image.
The last part is made a lot easier by noticing that the questions asks you to calculate the Jacobian at the point $(a,b) = (f_{1}(x,y), f_{2}(x,y)) \in A$, i.e. that $f^{-1}(a,b) = (x,y)$ (as we have proven now that $f$ is a bijection). So $(f^{-1} \circ f)(x,y) = (x,y)$, meaning that $d_{(x,y)}(f^{-1}\circ f) = \text{Id}$. But from the IFT we also know that
$$
d_{(a,b)}(f^{-1}) = (d_{(x,y)}f)^{-1},
$$
recalling that $(a,b)=f(x,y)$. I will leave the explicit calculations to you, but the final answer should be
$$ d_{(a,b)}(f^{-1}) = (d_{(x,y)}f)^{-1} =
\begin{bmatrix}
e^{-x}\cos y & -e^{-x}\sin y\\
e^{-x}\sin y & e^{-x}\cos y
\end{bmatrix}
$$
Best Answer
Yes, this can happen even in case of $n=1$, for example take $f(x) = x^3$. The inverse theorem makes a claim in the other direction, that is if the determinant of the Jacobian is not zero, then there is ... .