The curve is shown below.

Under the transformation, we note that
\[
x^{2/3} + y^{2/3} = u^{2/3} (\sin^2 v + \cos^2 v) = u^{2/3} = a^{2/3},
\]

and therefore the curve is mapped to the line $u= a$ in the $(u,v)$ plane. Where are the coordinate axes, $x = 0$ and $y = 0$ mapped? We can verify that since $y/x = \tan^3 v > 0$ in the first quadrant, then necessarily, $0 \leq v \leq \pi/2$. We can then verify that the line $x = 0$ corresponds to $v = \pi/2$ and the line $y = 0$ corresponds to $v = 0$. The Jacobian can be verified to be
\[
J = \pd{(x,y)}{(u,v)} = \begin{pmatrix}
\pd{x}{u} & \pd{x}{v} \\
\pd{y}{u} & \pd{y}{v}
\end{pmatrix}
= \begin{pmatrix}
\cos^3 v & -3u\cos^2 v \sin v \\
\sin^3 v & 3u\cos v \sin^2 v
\end{pmatrix},
\]

and we can verify that $|J| = 3u\cos^2 v \sin^2 v$. The computation of the area then follows:
\begin{align*}
A &= \iint_R \, \de{x} \, \de{y} \\
&= \int_{0}^{\pi/2} \int_0^a |J| \, \de{u} \, \de{v} \\
&= \int_{0}^{\pi/2} \int_0^a 3u\cos^2 v \sin^2 v \, \de{u} \, \de{v} \\
&= \frac{3a^2 \pi}{32}.
\end{align*}