Let $x^2-y^2 = u$ and $xy = v$
$x^2+y^2 = \sqrt{(x^2-y^2)^2 + 4 x^2 y^2} = \sqrt{u^2 + 4 v^2}$
$f(u,v) = v\sqrt{u^2 + 4 v^2}$
$I = \int \int f(x,y) dx dy = \int \int f(u,v)|J| du dv$
Now, $|J|^{-1} = \begin{vmatrix}
\frac{du}{dx} & \frac{du}{dy} \\
\frac{dv}{dx} & \frac{dv}{dy}
\end{vmatrix} \\ = \begin{vmatrix}
2x & -2y\\
y & x
\end{vmatrix} = 2x^2 + 2y^2 = 2 \sqrt{u^2 + 4 v^2}$
Thus, $I = \int \int v\sqrt{u^2 + 4 v^2}\ \frac{1}{2\sqrt{u^2 + 4 v^2}} du dv \\
= \frac{1}{2}\int \int v du dv\\
= \frac{1}{2}\int_{v=1}^{4} \int_{u=-3}^{3} v dudv\\
= \frac{1}{2}\int_{v=1}^{4} vu \bigg\rvert_{u= -3}^{3}dv \\
= \frac{1}{2}\int_{v=1}^{4} 6v dv \\
= \frac{6}{4} v^2\bigg\rvert_{1}^{4} \\
= \frac{6}{4}(4^2-1^2)= \frac{45}{2}$