Let us work in the "nice" situation where $X,Y,Z$ are smooth complex algebraic varieties, not necessarily compact. Assume that the fiber product $W:= X \times\_Z Y$ is also smooth. What can we say about the Picard group of $W$? 


More precisely, assume that $Pic(X)=Pic(Z)=0$.


1) Can we deduce that $Pic(W)=Pic(Y)$?


2) If 1) is not true in general, can we draw the conclusion if $Z$ is a point and thus $W=X \times Y$?


3) If 1) is not true in general, can we draw the conclusion if $Y \to Z$ is a finite étale cover?



Picard groups of (fiber) products
It can happen that $Pic(X) = Pic(Y) = Pic(Z) = 0$ but $Pic(W) \neq 0$!


For example, let $f: \mathcal{E} \to Z$ be a non-isotrivial family of elliptic curves, where $Z$ is a smooth rational curve. Then $Pic(\mathcal{E})$ is finitely generated, so by removing a finite number of (images of) sections of $f$ we obtain a surface $X \to Z$ such that $Pic(X) = 0$. 
Letting $Y = X$, I claim that $Pic(W) \neq 0$:


To see this, let $\Delta$ be the diagonal inside $W = X \times\_Z X$ so $L :=\mathcal{O}(\Delta) \in Pic(W)$. The restriction of $L$ to any fibre of the map $W \to Z$ is nonzero
since the cohomology class of the diagonal in $E \times E$, where $E$ is any elliptic curve, remains non-zero when restricted to $E' \times E'$ where $E' \subset E$ is any non-empty Zariski open subset. In particular, $L \neq 0$, so $Pic(W) \neq 0$.


3) is also false. 


For example, let $Y \to Z$ be a finite etale cover such that $Z$ is a smooth rational curve and the genus of $\bar{Y}$, the smooth compactification of $Y$, is at least $2$. Let $X$ also be a rational curve and take any morphism $X \to Z$ of degree $> 1$. Then $W$ is a smooth curve with a map to $Y$ of degree $> 1$. Since $g(\bar{Y}) > 1$, at least one component of $\bar{W}$ has genus $> g(\bar{Y})$ or it has more than one component, so the cokernel of the induced map $Pic(\bar{Y}) \to Pic(\bar{W})$ is uncountable. Since the kernel of the map $Pic(\bar{W}) \to Pic(W)$ is finitely generated, it follows that the the map $Pic(Y) \to Pic(W)$ cannot be surjective. 


