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 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 fiber bundle?

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

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?