An exercise 2.3 from Harris's AG: A First Course This is an exercise 2.3 from Harris's AG: A First Course:

Using the result of the exercise 2.2 ($A(\mathbb{A}^2 \setminus \{0\}) = K[x_1, x_2]$) show that $X = \mathbb{A}^n \setminus \{0\}$, $n > 1$, [as a quasi-affine variety] is not isomorphic [as in biregular] to any affine variety.

I have serious issues with this exercise.
I'm going to use the fact that regular mappings are Zariski continuous. I am also assuming that the underlying field $K$ is infinite.
Consider a regular isomorphism $f: X \to Y$ where $Y$ is an affine subvariety of $\mathbb{A}^m$ cut out by polynomials $h_i$, $i \in I$. By (a slight generalization of) ex. 2.2 we know that the components of $f$ are exactly restrictions of polynomials to $X$, we'll denote the extension of $f$ to the whole $\mathbb{A}^n$ as $\hat{f}$. But then $h_i \circ \hat{f} \vert_{\mathbb{A}^1}$ has infinite number of zeros, which means that the image of $0 \in \mathbb{A}^n$ lies in $Y$ as well.
Consider now a line $L \subset \mathbb{A}^n$ passing through both preimages of $\hat{f}(0)$. It is now sufficient to prove that $\hat{f} \vert_{X \cap L}$ is not an isomorphism onto its image.
I don't really know how to proceed. It is clear that $X \cap L$ is isomorphic to a hyperbola $xy = 1$, and thus we can reduce the problem to the one about affine varieties. However, I don't know how to prove that they are not isomorphic without using Euclidean topology (and thus assuming that $K = \mathbb{R}$ or $K = \mathbb{C}$. I guess I could look ahead and use the definition of a singularity, but it's so far ahead in the textbook it would be cheating. I can't think of a nice Zariski closed set that $\hat{f}$ would not pull back onto a Zariski closed set either, and I'm pretty sure one doesn't exist. So I'm left with having to study coordinate rings of the hyperbola and its image, and I don't know how to reason about the latter.
Am I on the wrong track with this exercise? I suspect that I am, yet I don't know any better way to go about it. Is there a hint I could use?
 A: Your argument begins by showing that an isomorphism $f:X \cong Y$ with $Y$ affine
has to extend to a morphism $\hat{f}:\mathbb A^n \to Y.$
Next, your idea is to choose $x \in X$ such that $\hat{f}(x) = \hat{f}(0)$, and
take the line $L$ through $x$ and $0$.  
This seems like a good strategy.  Perhaps you could use the following to go further: since $f$ is an isomorphism and $L \setminus \{0\}$ is closed in $X$,
its image $f(L\setminus\{0\})$ is closed in $Y$.   But general topology shows that since $L$ is the closure of $L\setminus \{0\}$, there is an inclusion $\hat{f}(L) \subset$ closure of $f(L\setminus\{0\})$.   Thus in fact $\hat{f}(L) = f(L\setminus\{0\}).$  
Now $f(L\setminus \{0\})$ is just an isomorphic copy of $L\setminus \{0\}$,
so you are reduced to the following question: is there a morphism from $L$ to $L\setminus \{0\}$? 
Further remarks: If we compose with $f^{-1}$, then we can eliminate some notation, and then what you are trying to show is that the identity morphism 
$X \to X$
does not extend to a morphism $\mathbb A^n \to X$.  To prove this, you are restricting to a line, i.e. a copy of $\mathbb A^1$, passing through $0$,
and showing that the identity $\mathbb A^1 \setminus \{0\} \to \mathbb A^1\setminus \{0\}$ doesn't extend to a morphism $\mathbb A^1 \to \mathbb A^1\setminus \{0\}.$ (So actually you don't need to choose $L$ containing $x$ in the preiamge of $\hat{f}(0)$; as long as it passes through $0$ it should lead to a contradiction.)
