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No. As an extreme example, suppose that $g$ is the identity (which is etale everywhere), and that $f$ is not etale at some point. Then the fibre product is just $f$ again.

But in fact, this is essentially the general case. If $g$ is etale (or smooth) at a point, then it is etale (resp. smooth) in a n.h. of that point, so we may replace $Z$ by the n.h. and so assume that $g$ is etale everywhere. Then if $f$ is not etale (or smooth) at a point $y \in Y$, the product will not be etale in a n.h. of $y \times Z.$

(Imagine that $Y$ was e.g. a nodal curve with a node at $y$, and that $Z$ is a smooth curve. (Here $X$ is Spec of the ground field.) Then $Y\times Z$ is the product of a nodal curve and a smooth curve, which just looks like a cylinder over the nodal curve; it is singular all along the "cylinder" over the node.)

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No. As an extreme example, suppose that $g$ is the identity (which is etale everywhere), and that $f$ is not etale at some point. Then the fibre product is just $f$ again.

But in fact, this is essentially the general case. If $g$ is etale (or smooth) at a point, then it is etale in n.h. of that point, so we may replace $Z$ by the n.h. and so assume that $g$ is etale everywhere. Then if $f$ is not etale (or smooth) at a point $y \in Y$, the product will not be etale in a n.h. of $y \times Z.$

(Imagine that $Y$ was e.g. a nodal curve with a node at $y$, and that $Z$ is a smooth curve. (Here $X$ is Spec of the ground field.) Then $Y\times Z$ is the product of a nodal curve and a smooth curve, which just looks like a cylinder over the nodal curve; it is singular all along the "cylinder" over the node.)