Differentiate tan⁻¹(x√1-x²) with respect to sin⁻¹(2x√1-x²)

Let y = tan⁻¹(x√1-x²)
x = sinθ
y = tan⁻¹(sinθ√1-sin²θ) = tan⁻¹(sinθcosθ) = tan⁻¹(½sin2θ)
Differentiating with respect to θ,
dy/dθ = 1/(1+(½sin2θ)²) * ½(2cos2θ) = cos2θ/(1+¼sin²2θ)
Let z = sin⁻¹(2x√1-x²) = sin⁻¹(2sinθ√1-sin²θ) = sin⁻¹(2sinθcosθ) = sin⁻¹(sin2θ) = 2θ
Differentiating with respect to θ,
dz/dθ = 2
Now dy/dz = (dy/dθ)/(dz/dθ) = (cos2θ/(1+¼sin²2θ))/2 = cos2θ/(2+½sin²2θ)