Let f(θ) = sin(tan⁻¹(sinθ√cos2θ)), where -π/4 < θ < π/4. Then the value of d/d(tanθ)(f(θ)) is?

Given f(θ) = sin(tan⁻¹(sinθ√cos2θ))
Let y = tan⁻¹(sinθ√cos2θ)
Then, tany = sinθ√cos2θ
sin2y = 2tany/(1+tan²y) = 2sinθ√cos2θ/(1+sin²θcos2θ)
Let u = sinθ√cos2θ
Then f(θ) = siny
df/dθ = cosy(dy/dθ)
d(tany)/dθ = d(sinθ√cos2θ)/dθ
Sec²y(dy/dθ) = cosθ√cos2θ + sinθ(-2sin2θ/2√cos2θ)
Sec²y(dy/dθ) = cosθ√cos2θ - sin²θ(sin2θ/√cos2θ)
d/d(tanθ)(f(θ)) = (df/dθ)/(d(tanθ)/dθ) = (cosy(dy/dθ))/sec²θ
Let θ = 0
f(0) = sin(tan⁻¹(0)) = 0
d/d(tanθ)(f(θ)) = (df/dθ)/(d(tanθ)/dθ) = (cosy * (cosθ√cos2θ - sin²θ(sin2θ/√cos2θ))/sec²θ)
At θ = 0, cosy = 1, dy/dθ = 1
d/d(tanθ)(f(θ)) = 1/1 = 1