Answer
$\dfrac{3e-7}{6}$
Work Step by Step
$ \iiint_E e^{z/y} dV=\int_{0}^1 \int_{y}^{1} \int_{0}^{xy}e^{z/y} dz dx dy\\= \int_{0}^{1} \int_{y}^{1} [ye^{z/y}]_{0}^{xy} dx dy\\=\int_{0}^1 [ye^{x}-xy]_y^1 dy\\=\int_{0}^1 [(1-0)e^{x}-x(1-0)] dy\\
= \int_{0}^{1} [yx+y^2-yx+y^2] dy dx\\= \int_0^{1}[(e-1)y-ye^y+y^2] dy\\=[(e-1)(y^2/2)-ye^y+(y^3/3)]_0^1 dy\\\\=\dfrac{3e-7}{6}$
