Let y = sin^-1 x/√(1 - x² )

=> y√(1 - x² ) = sin^-1 x

Differentiate w.r.t. x, we get

       d{y√(1 - x² )}/dx = d(sin^-1 x)/dx

=> √(1 - x² )*(dy/dx) + y * d{√(1 - x² )}/dx = 1/√(1 - x² )

=> √(1 - x² )*(dy/dx) + y * (-2x)/{2*√(1 - x² )} = 1/√(1 - x² )

=> [√(1 - x² )*(dy/dx) - xy/{*√(1 - x² )}]*√(1 - x² ) = 1

=> (1 - x² )*(dy/dx) - xy = 1

Again differentiate w.r.t. x, we get,

      d{(1 - x² )*(dy/dx) - xy}/dx = d1/dx

=> (1 - x² )*(d² y/dx² ) - 2x*(dy/dx) - x*(dy/dx) - y = 0

=> (1 - x² )*(d² y/dx² ) - 3x*(dy/dx) - y = 0

Hence proved.