Here is an elemantary example:


Define $f:S^1\times \mathbb{C}\rightarrow\mathbb{C}$ by $f(\zeta, z)=\zeta\cdot z^n$, where $n\geq 2$ is an integer, then $f$ is a smooth map, every fiber of $f$ is a smooth submanifold of $S^1\times\mathbb{C}$ and is diffeomorphic to $S^1$. However, $(S^1\times \mathbb{C}, \mathbb{C}, f)$ is not a differentialble fiber bundle.


**Question**: Is there any similar example in algebraic geometry?


More Precisely: 


$f:X\rightarrow \mathbb{C}P^1$ is a smooth morphism, where $X$ is a smooth projective algebraic variety over $\mathbb{C}$. Every fiber of $f$ is a smooth subvariety of $X$. 


Does it imply that $(X, \mathbb{C}P^1, f)$ is a differentiable fiber bundle? 


What will happen if we replace $\mathbb{C}P^1$ by a smooth variety $Y$ in general?



The smoothness of fiber and fiber bundle
By definition, a morphism $f \colon X \to Y$ between smooth projective varieties is smooth if $f$ is flat and all fibres are smooth (in the scheme-theoretical sense). In particular, $f$ is a smooth submersion when it is considered as a differentiable map between real manifolds.


Then the answer to your question is *yes*, assuming that the fibres are connected. This remains true also replacing $\mathbb{P}^1$ with any smooth projective variety $Y$.


In fact, by a result due to Ehresmann (1951), if $ƒ \colon M \to N$ is a surjective submersion with $M$ and $N$ differentiable manifolds such that the preimage $ƒ^{-1}(x)$ is compact and connected for all $x \in N$, then $ƒ$ is differentiably locally trivial and admits a compatible fiber bundle structure. See <http://en.wikipedia.org/wiki/Fiber_bundle>. 


In particular, all fibres of $f$ are diffeomorphic.


**Remark 1.** If one only requires that the fibres of $f \colon X \to Y$ are smooth in the set-theoretical sense, and not in the scheme-theoretical one (i.e. one allows multiple fibres) then in general $f$ is *not* a differentiable fibre bundle, see Jason Starr's comment. 


**Remark 2.** Even if $f \colon X \to Y$ is differentiably locally trivial, in general it is *not* analitycally locally trivial. In fact, by a celebrated theorem of Grauert and Fischer, this happens if and only if all the fibres of $f$ are biholomorphic.
See for instance [Barth-Peters-Van de Ven, Compact Complex Surfaces, Chapter I].


