$$
\begin{split}
\frac{dy}{dx} & = \frac{dy}{d\log y} \frac{d\log y}{dx} \\
& = \left(\frac{d\log y}{dy}\right)^{-1} \frac{d\log y}{dx} \\
& = y \cdot \frac{d\log y}{dx}
\end{split}
$$
In practice this means that the marginal effect of $x$ on $y$ *depends on the value of $y$, so you'd have to pick reference values of $x$ and $z$ (typically the mean values) to do the calculation.
Alternatively, you can frame the marginal effect on a log-response as a proportional change. If you compute the marginal effect of $x$ on $\log y$ and exponentiate it, you can frame it as a proportional effect (e.g. "a one-unit change in $x$ is associated with a ()-fold change in $y$")