When is a birational bijection étale? So this question is probably not "research level", although, for what it is worth, it is coming up in a research paper I am presently writing.
Let $X,Y$ be irreducible affine varieties over $\mathbb{C}$.  Suppose $f:X\to Y$ is a bijective birational morphism.  Assume that $X$ is normal, but $Y$ is not and that both $X,Y$ are singular (not smooth).
Question 1: Is $f$ étale?
I know that $f$ is the normalization map, and that if $Y$ is normal then $f$ is an isomorphism in this setting.  I also know that the Grothendeick classes of $X$ and $Y$ are equal (induced by $f$). Also, at corresponding smooth points there is a local analytic isomorphism and so at such points the map is étale.
I suspect the answer to this question is no, and that there are easy counter-examples (that I have probably seen before and am just forgetting now).
So let me ask a second question that might be more useful to me than the expected counter-example.
Question 2: Are there easy to check (think computational) conditions on $X$, $Y$, or $f$ that will guarantee that $f$ is étale?
 A: This map is étale if and only if $Y$ is normal.
You mentioned the proof of one direction. In the other, probably the most straightforward proof uses the invariance of normality under étale morphisms of local rings. If $f: X \to Y$ is étale, this result implies $X$ is normal at $x$ if and only if $Y$ is normal at $f(x)$. Since $f$ is bijective, and $X$ is normal, this implies $Y$ is normal.
Some alternate proofs I came up with while looking for the reference for that one:
(1) Using this result from the stacks project, we can see that, because $f$ is the normalization map, it is finite, and if it is étale, it is finite, bijective, and étale, so by this lemma is Zariski-locally an isomorphism, hence is an isomorphism, so $Y$ is normal.
(2) Assume $f$ is étale, and note that $ X\times_Y X \to X$ is the base change of a bijective birational map $f: X \to Y$ by an étale morphism $f: X \to Y$, hence is also bijective and birational. Similarly $X\times_Y X$ is normal as it has an étale map to a normal variety. So, by the case you already did $ X\times_Y X \to X$ is an isomorphism. Since the base change of $f$ is an isomorphism, $f$ is an isomorphism.
