= Solution
For a one-to-one continuously differentiable change of variables with nonzero <Jacobian determinant>, the <change of variables formula> rule is
$$
\iint_D F(x,y)\,dx\,dy
=\iint_E F(x(u,v),y(u,v))
\left|\det\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv,
$$
where $E$ is the transformed region and the inverse map covers $D$ once. The absolute value is required even if the coordinate map reverses orientation.
Here $x,y$ are positive. For $u=y/x$, $v=xy$, the inverse is $x=\sqrt{v/u}$ and $y=\sqrt{uv}$. On the first portion of $D$, the bounds give $v\ge1$, $u\le4$ and $v\le u$ because $x\le1$. Together these give $1\le v\le u\le4$; conversely these inequalities imply $1/2\le x\le1$ and the original bounds. On the second portion, they give $u\ge1$, $v\le4$ and $v\ge u$, hence $1\le u\le v\le4$, with the converse following similarly. The two triangles combine into the rectangle \b[$E=\{1\le u\le4,\ 1\le v\le4\}$], sharing only the diagonal corresponding to $x=1$.
The Jacobian is
$$
\det\frac{\partial(u,v)}{\partial(x,y)}
=\det\begin{pmatrix}-y/x^2&1/x\\y&x\end{pmatrix}=-\frac{2y}{x}=-2u,
\qquad
\left|\det\frac{\partial(x,y)}{\partial(u,v)}\right|=\frac1{2u}.
$$
Also $xy^3=uv^2$ and $x^2+y^2=v(1+u^2)/u$. The transformed integral consequently factors:
$$
\iint_D\frac{4xy^3}{x^2+y^2}\,dx\,dy
=\int_1^4\int_1^4\frac{2uv}{1+u^2}\,dv\,du
=\left[\frac{v^2}{2}\right]_1^4
\left[\log(1+u^2)\right]_1^4.
$$
Therefore
$$
\boxed{\iint_D\frac{4xy^3}{x^2+y^2}\,dx\,dy=\frac{15}{2}\log\frac{17}{2}.}
$$
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