A question on smooth morphisms and 'pointwise' smooth morphisms Let $X$ be a scheme, $x\in X$ a point and $f\colon \operatorname{Spec}(k(x))\to X$ the canonical morphism.

Is $f$ always a smooth morphism?

Now suppose $g\colon X\to Y$ is a scheme over some other scheme $Y$.

What is the relation between the following two statements?
  
  
*
  
*For each $x\in X$, the composition $\operatorname{Spec}(k(x))\to X \to Y$ is a smooth morphism.
  
*$g$ is a smooth morphism.
  

Is it true that (2.) $\Leftrightarrow$ (1.)+($g$ is flat)+(something)?
 A: The morphism $\mathrm{Spec}(k(x))\to X$ need not be locally of finite type, so it in general has no chance to be smooth. For example, take $X=\mathbf{A}_k^1$ for a field $k$ and $x$ the generic point. The morphism corresponds to the inclusion $k[T]\hookrightarrow k(T)$, and this is not of finite type, hence not smooth. At least if $X$ is a $k$-scheme for some field $k$, and is locally of finite type over $k$, then for $x\in X$ a closed point, the morphism $\mathrm{Spec}(k(x))\to X$ is locally of finite type, so there is a chance for smoothness, but even here it's not generally true. Indeed, for any field $k$ which is not algebraically closed and any irreducible polynomial $f(T)\in k[T]$ of positive degree corresponding to the (closed) point $x\in X:=\mathbf{A}_k^1=\mathrm{Spec}(k[T])$, $k(x)=k[T]/(f(T))$ and the morphism in question corresponds to the surjection $k[T]\to k[T]/(f(T))$, and hence is a closed immersion. Smooth morphisms are flat, and this is never flat because $k[T]/(f(T))$ is a torsion $k[T]$-algebra. 
To see that (2) need not imply (1), take the situation in the previous paragraph but with $k$ imperfect of characteristic $p>0$ and $f(T)$ of the form $T^p-\alpha$ with $\alpha\in k-k^p$. The morphism $X\to Y=\mathrm{Spec}(k)$ is smooth (basically by definition), but the field $k(x)=k[T]/(T^p-\alpha)$ is a finite purely inseparable extension of $k$, so the composite $\mathrm{Spec}(k(x))\to X\to\mathrm{Spec}(k)$ is definitely not smooth (e.g. the source is not geometrically reduced over $k$ as can be seen by base changing along $k\to k(x)$).
I don't think the condition in (1) is especially sensible for the reason I give above. If it holds, then $\mathrm{Spec}(k(x))\to X$ is locally of finite type for all $x\in X$, and, e.g. if $X$ is a $k$-scheme of finite type, this will never hold unless $X$ is zero-dimensional. However, if $X$ is zero-dimensional and of finite type over $k$, then if each composite $\mathrm{Spec}(k(x))\to X\to Y=\mathrm{Spec}(k)$ is smooth (equivalently each $k(x)$ is separable over $k$ for each $x\in X$), then this does indeed imply that $X$ is $k$-smooth (equivalently étale).
EDIT: Also, the term "fiberwise" smooth morphisms has nothing to do with your question. The scheme-theoretic fiber of a morphism $f:X\to Y$ over a point $y\in Y$ is the $k(y)$-scheme $X_y=X\times_Y\mathrm{Spec}(k(y))$. If $f:X\to Y$ is flat, locally of finite presentation, and each scheme-theoretic fiber is smooth over its respective residue field, then $f$ is smooth. This is either a theorem or the definition of smoothness, depending on what definition you choose. 
EDIT 2: I would just add that "pointwise" smooth most reasonably can be interpreted as "smooth at a point." A morphism $f:X\to Y$ is smooth at a point $x\in X$ if there are opens $x\in U\subseteq X$ and $f(x)\in V\subseteq Y$ with $f(U)\subseteq V$ and the morphism $U\to V$ smooth (or we could even require $U$ and $V$ to be affine and the ring map $\mathscr{O}_Y(V)\to\mathscr{O}_X(U)$ to be smooth). This is the way the Stacks Project defines smoothness (first defining smoothness of a ring map). This definition has the advantage that the locus of points in $X$ where $f:X\to Y$ is smooth is obviously open (every point in the open $U$ is a smooth point). 
