The Jacobian for this transformation is
[tex]J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}[/tex]
with determinant [tex]|J| = 12[/tex], hence the area element becomes
[tex]dA = dx\,dy = 12 \, du\,dv[/tex]
Then the integral becomes
[tex]\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv[/tex]
where [tex]R'[/tex] is the unit circle,
[tex]\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1[/tex]
so that
[tex]\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv[/tex]
Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.
[tex]\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}[/tex]
Then
[tex]\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}[/tex]
