$x\frac{dy}{dx}=\sqrt{1-y^2}$
$\frac{dy}{dx}=\frac{\sqrt{1-y^2}}{x}$
Note this is a separable function, rearrange:
$\int{\frac{1}{\sqrt{1-y^2}}dy}=\int{\frac{1}{x}dx}$ where $x≠0, y≠±1$
Integrate both side:
$LHS: \int{\frac{1}{\sqrt{1-y^2}}dy}=\arcsin{y}$
$RHS: \int{\frac{1}{x}dx}=\ln{|x|}+C$
$arcsin(y)=\ln{|x|}+C$
∴General Solution is the following:
$y=\sin{(\ln{|x|}+C)}$ where $x≠0, y≠±1$
