Let $k$ be a field and let $A = k[x_1,x_2,\ldots]$. Note that $A$ is not of finite type over $k$.
Is $\operatorname{Spec} A\to \operatorname{Spec} k$ a smooth morphism of schemes?
I think it is formally smooth, but I'm not sure if it is smooth because it is not of finite type.
Smooth morphisms are locally of finite presentation by definition. Since $k[x_1,x_2,\dotsc]$ is not finitely generated (hence, not finitely presented) as a $k$-algebra, the morphism is not smooth.
But it is formally smooth: If $B$ is a commutative $k$-algebra and $I \subseteq B$ is a nilpotent ideal (in fact, every ideal works here), then every $k$-algebra homomorphism $k[x_1,x_2,\dotsc] \to B/I$ lifts to a $k$-algebra homomorphism $k[x_1,x_2,\dotsc] \to B$. This follows directly from the universal property of polynomial algebras (and the axiom of choice).