Answer
$$e-2$$
Work Step by Step
$$\iint_{R} dA= \int_{0}^{1} \int_{0}^{y^2} 3 y^3 e^{xy} dx dy\\= \int_{0}^{1} [3y^2 e^{xy}]_{0}^{y^2} dy \\= \int_0^{1} 3y^2(e^{y^3}-1) dy $$
Suppose $y^3=u$ and $du= 3y^2dy$
Now, $$\iint_{R} dA=\int_0^1 e^{u}-1 du \\ =[e^u-u]_0^1 \\=e^1-1-1-0 \\=e-2$$
