y'=[f(sin^2 x)]'+[f(cos^2 x)]'
=f'(sin^2 x)*(sin^2 x)'+f'(cos^2 x)*(cos^2 x)'
=f'(sin^2 x)*2sinx*(sinx)'+f'(cos^2 x)*2cosx*(cosx)'
=f'(sin^2 x)*2sinx*cosx+f'(cos^2 x)*2cosx*(-sinx)
=sin2xf'(sin^2 x)-sin2xf'(cos^2 x)
=sin2x*[f'(sin^2 x)-f'(cos^2 x)]
所以dy/dx=sin2x*[f'(sin^2 x)-f'(cos^2 x)]