= Solution
First take $a>0$ and $|x|<a$, and put $u=\arcsin(x/a)$, $q=\sqrt{a^2-x^2}$. Then $u'=1/q$. The <chain rule> gives
$$
y'=\frac{2u}{q},\qquad y''=\frac2{q^2}+\frac{2xu}{q^3}.
$$
Consequently
$$
(a^2-x^2)y''-xy'=q^2\left(\frac2{q^2}+\frac{2xu}{q^3}\right)-\frac{2xu}{q}=2,
$$
which proves the identity on the open domain of differentiability. The squared inverse sine is unchanged when $a$ is replaced by $-a$, so the identity also holds for negative $a$ with $|x|<|a|$. The value $a=0$ is excluded.
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