The integrating factor of the differential equation.

The given differential equation is:

$\left(1-y^{2}\right) \frac{d x}{d y}+y x=a y$

$\Rightarrow \frac{d y}{d x}+\frac{y x}{1-y^{2}}=\frac{a y}{1-y^{2}}$

This is a linear differential equation of the form:

$\frac{d x}{d y}+p y=Q$ (where $p=\frac{y}{1-y^{2}}$ and $Q=\frac{a y}{1-y^{2}}$ )

The integrating factor (I.F) is given by the relation,

$e^{\int p d x}$

$\therefore$ I.F $=e^{\int \mu d y}=e^{\int \frac{y}{1-y^{2} d y}}=e^{\frac{-1}{2} \log \left(1-y^{2}\right)}=e^{\log \left[\frac{1}{\sqrt{1-y^{2}}}\right]}=\frac{1}{\sqrt{1-y^{2}}}$

Hence, the correct answer is D.