= Solution
For $x>0$, differentiation of $u=y/x$ gives
$$
u'=\frac{y'}x-\frac y{x^2},\qquad
u''=\frac{y''}x-\frac{2y'}{x^2}+\frac{2y}{x^3},\qquad
u''+\frac2xu'-u=\frac{y''-y}{x}.
$$
Thus the <radial Laplacian> equation reduces to
$$
\boxed{y''-y=xg(x),\qquad f(x)=xg(x)}.
$$
Boundedness of $u$ at zero implies $y(0)=0$. Conversely, a regular reduced solution with $y(0)=0$ gives finite $u(0)=y'(0)$.
There is an important qualification to the reduction in the printed question: bounded $u$ at infinity requires $y=O(x)$, not bounded $y$. The example in part (ii) actually has $y\sim-x$, so it cannot be the bounded solution described in part (a). The differential equation and origin condition reduce as asserted, and the same <Green's function> can still be used with the correct growth condition. This is the <radial reduction and growth at infinity> distinction.
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