The result you ask about is called Hadamard's global inverse function theorem or sometimes Hadamard-Cacciopoli theorem. Googling these keywords reveals an entire industry of such local invertibility + something implies global bijectivity results.
Unfortunately, I was unable to find an accessible proof of this result. Among several sources I looked at, by far the best bet seems to be the presentation in Section 6.2 of the beautiful book by S.G. Krantz and H.R. Parks, The implicit function theorem: history, theory, and applications, Birkhäuser, 2002. The proof given there is essentially self-contained and doesn't assume much knowledge on the reader's side. Nevertheless, I should point out that the title of Chapter 6 is Advanced implicit function theorems, so it's definitely not for the faint-hearted.
In fact, a more general result is the following, also due to Jacques Hadamard. It is a bit, but not very much, harder to prove than the result you ask about.
If you don't know what a manifold is, simply replace $M_1$ and $M_2$ by $\mathbb{R}^n$ in the theorem below, and you obtain the result you're asking about — for $\mathbb{R}^n$ condition 3. is satisfied and condition 1. translates precisely to the condition $\lim\limits_{|x| \to \infty} |f(x)| = \infty$ your tutor told you.
Theorem (Hadamard)
Let $M_1, M_2$ be smooth and connected $n$-dimensional manifolds. Suppose $f: M_1 \to M_2$ is a $C^1$-function such that
- $f$ is proper
- The Jacobian of $f$ is everywhere invertible
- $M_2$ is simply connected.
Then $f$ is a homeomorphism (hence globally bijective).
So, as I said, this theorem is not trivial at all and both this and the result you're interested in can be found in the book I mentioned above. Quite a bit of googling didn't yield a simple(r) proof of the theorem you ask about, but as you have the key-words now, maybe you find something that suits you.
Added: I should have mentioned the better known Cartan-Hadamard theorem which is closely related but seems a bit more geometric in its nature.
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 $2\pi$.
Best Answer
Let me give you a sketch of a possible argument: First show $f$ is injective. You can do this by definition.
Next show $f$ is surjective. I would recommend showing, that the image of $f$, i.e. $$f(\mathbb{R}^n)\subseteq \mathbb{R}^n,$$ is open, closed and not empty. Since $\mathbb{R}^n$ is connected, we would then have $$f(\mathbb{R}^n)= \mathbb{R}^n,$$ which is surjectivity. To show that the image is open, you can use the 'local' inverse function theorem. The closedness can be done via sequences (Cauchy criterion for convergence) and the assumed inequality on $f$. This shows you, that $f$ is bijective, hence can be inverted.
The regularity of the inverse function follows again from the 'local' inverse function theorem, since differentiability and continuity are local properties.