Answer
$$y'=4(x+(x+\sin^2x)^3)^3(1+3(x+\sin^2x)^2(1+\sin2x))$$
Work Step by Step
$$y'=\left((x+(x+\sin^2x)^3)^4\right)'=
4(x+(x+\sin^2x)^3)^3\cdot((x+(x+\sin^2x)^3)'=
4(x+(x+\sin^2x)^3)^3\cdot(1+3(x+\sin^2x)^2\cdot(x+\sin^2x)')=
4(x+(x+\sin^2x)^3)^3(1+3(x+\sin^2x)^2(1+2\sin x\cdot(\sin x)'))=
4(x+(x+\sin^2x)^3)^3(1+3(x+\sin^2x)^2(1+2\sin x\cos x))=
4(x+(x+\sin^2x)^3)^3(1+3(x+\sin^2x)^2(1+\sin2x))$$
