Intuition for étale morphisms

Instead of answering your question with general results easily obtainable from the literature, online or traditional, I'll give you a few morphisms.
Deciding whether they are are étale may contribute to developing your intuition.
(Of course I'll gladly help you or anybody else if you had any problem with these morphisms)

a) $\mathbb A^1_\mathbb C\to \operatorname {Spec}(\mathbb C[X,Y]/(Y^2-X^3)): t\mapsto (t^2,t^3)$
b) $\mathbb A^1_\mathbb C\to \operatorname {Spec}(\mathbb C[X,Y]/(Y^2-X^2-X^3)): t\mapsto (t^2-1,t^3-t)$
c) $\mathbb A^2_\mathbb C\to \mathbb A^2_\mathbb C: (x,y)\mapsto (x,xy)$
d) $\operatorname {Spec}\mathbb C[T]\to \operatorname {Spec}\mathbb C[T^2,T^3]$
e) $\operatorname {Spec}\mathbb Q[T]/(T^2-4)\to \operatorname {Spec}\mathbb Q$
f) $\operatorname {Spec}\mathbb Q[T]/(T^2+4)\to \operatorname {Spec}\mathbb Q$
g) $\operatorname {Spec}\mathbb Q[T]/(T^2)\to \operatorname {Spec}\mathbb Q$
h) $\operatorname {Spec}\mathbb F_9\to \operatorname {Spec}\mathbb F_3$

Edit (one day later) : Two useful theorems and how they settle the question of étaleness of the above morphisms

Theorem 1 Given a field $k$ and a $k$-algebra $A$, the morphism $\operatorname {Spec}(A)\to \operatorname {Spec}(k) $ is étale iff $A$ is isomorphic as a $k$-algebra to a finite product $A\cong K_1\times...\times K_n$ of finite separable field extensions $K_i/k$.
Remark In the étale case, $A$ must be reduced ( i.e. $\operatorname {Nil}(A)=0$ )
Example Every finite Galois extension $K/k$ gives rise to an étale morphism $\operatorname {Spec}(K)\to \operatorname {Spec}(k) $. This is the kernel of Grothendieck's famous geometrization of Galois theory
Illustration The morphisms e), f), h) are étale but g) is not because $\operatorname {Spec}\mathbb Q[T]/(T^2)$ is not reduced.

Theorem 2 A morphism of schemes $f:X\to Y$ is étale iff it is flat and unramified.
Illustration The morphisms a) , b), c) and d) are not étale because they are not flat
For the sake of completeness let me mention that a), c) and d) are ramified but that b) is unramified.
Let me also mention that a) and d) are two different presentations of the same morphism.

For four of Georges examples I can give a general hint. If $X\to Spec \ k$ is étale where $k$ is a field, then $X$ must just be a disjoint union $\coprod Spec \ L_i$ where each $L_i$ is a finite separable field extension of $k$.

Proving this fact is a good exercise and it goes back to the analogy people have been making. A "covering space" of a point should just topologically be a discrete set of points, but this is algebraic geometry so there should be algebraic information as well. The algebraic information is that the field extensions are all separable. This has to do with the fact that $X$ being smooth implies it is "geometrically reduced" and hence you see that we can't pick up nilpotents when base changing.

Now it should be pretty easy to do (e)-(h). Unfortunately, all the base fields there are perfect, so we don't have weird unnecessary complication. I'll throw in

i) $\displaystyle Spec \left(\frac{\mathbb{F}_p(t)[x]}{(x^p-t)}\right)\to Spec \ \mathbb{F}_p(t)$

To elaborate on Gunnar's comment: covering spaces provide good intuition, at least in characteristic zero. In positive characteristic you have to keep the Frobenius in mind at all times. Flatness expresses the fact that all the fibers of a connected covering have the same cardinality. Smoothness means that etale morphisms are surjections, hence isomorphisms (this is the relative dimension zero part), on tangent spaces, so in the complex topology they are local homeomorphisms by the inverse function theorem.