(As you have also stated the application you have in mind I assume that $X$ has everywhere the same dimension as $Y$ with no embedded components.)

We have that $H^i_{et}(X,\mathbf{G}_m) = H^i_{et}(Y,f_\ast\mathbf{G}_m)$ as $f$ is finite. If $f$ is also flat we have a norm map $f_\ast\mathbf{G}_m \to \mathbf{G}_m$ whose composite with the inclusion $\mathbf{G}_m \to f_\ast\mathbf{G}_m$ is the $d$'th power which gives what you want. In the general case you still have a norm map by first noting that $f$ is flat in codimension $1$ (this comes from the condition I added) and you can take the norm there which then will land in $\mathbf{G}_m$ as $Y$ is normal.

(As you have also stated the application you have in mind I assume that $X$ has everywhere the same dimension as $Y$ with no embedded components.)

We have that $H^i_{et}(X,\mathbf{G}_m) = H^i_{et}(Y,f_\ast\mathbf{G}_m)$ as $f$ is finite (this is because $f_\ast$ is exact, see Cor. II:3.5 of Milne: Étale cohomology). If $f$ is also flat we have a norm map $f_\ast\mathbf{G}_m \to \mathbf{G}_m$ whose composite with the inclusion $\mathbf{G}_m \to f_\ast\mathbf{G}_m$ is the $d$'th power which gives what you want. In the general case you still have a norm map by first noting that $f$ is flat in codimension $1$ (this comes from the condition I added) and you can take the norm there which then will land in $\mathbf{G}_m$ as $Y$ is normal.