Example of a non-smooth $Y\times_X Z$ over $Spec(K)$? Let $K$ be a field and $X\to Spec(K)$ and $Y\to  Spec(K)$ and $Z\to Spec(K)$ three smooth morphisms of schemes.
I have no intuition of relative smoothness. What is an example of such schemes and (not necessary smooth) morphisms $Y\to X$ and $Z\to X$ such that $Y\times_X Z$ is not smooth over $Spec(K)$?
Are there conditions (apart from ''it is an isomorphism'') on $X \to Spec(K)$ such that there are no examples?
 A: Here is an example where $Y\times_X Z$ is not smooth over $Spec(K) $.
Take $X=Spec(K[T])=\mathbb A^1_K, Y =Spec(K)=\lbrace *\rbrace $, and for $Z$ take the surface  $Z=V(Y^2-X^3-T)\subset Spec(K[X,Y,T])=\mathbb  A^3_K$.
The morphisms are $Z\to X:(x,y,t)\mapsto t$ and $Y\to X:*\mapsto 0 $.
The three schemes $X,Y,Z$ are smooth over $K$.
However the fiber product  $Y\times_X Z$ is the cusp $Y^2-X^3=0$ in the plane $Spec(K[X,Y])$ and is thus not smooth over $K$, just as you required.
A smoothness criterion
No reasonable condition on $X$ alone will ensure that $Y\times_X Z$ is smooth: after all in the example above $X=\mathbb A^1_K$ and it is impossible to conceive of a smoother, better behaved $K$-scheme than the affine line!
The correct condition you want is that the morphism $Y\to X$ be smooth and that $Z$ be smooth over  $K$.
Then the morphism $Y\times_X Z\to Z $ will be smooth too because smoothness is preserved by base change.
And the morphism $f: Y\times_X Z\to Spec(K) $ will be smooth,   just as you wished: indeed the composition of smooth morphisms is smooth and this morphism 
$f$ is the composition of the smooth morphisms  $Y\times_X Z\to Z $ and $Z\to Spec(K)$.
Notice carefully that you don't have to require that $X$ nor $Y$ be smooth over $K$ for the criterion to apply!
A criterion for the criterion
Yes, but how do we ensure that the morphism $f:Y\to X$ is smooth?
Answer: a necessary and sufficient condition for $f$ to be smooth is that $f$ be flat and that all fibers $f^{-1}(y)\subset X$ of closed points $y\in Y$ be smooth over $K$.
