y = sin^2 (3x-2) : find dy/dx?

Chain rule and double-angle formula.
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y = [ sin (3x - 2) ]²
dy/dx = 2 [ sin (3x - 2) ] [ 3 cos (3x - 2) ]
dy/dx = 6 [ sin (3x - 2) ] [ cos (3x - 2) ]
dy/dx = 3 [ 2 ] [ sin (3x - 2) ] [ cos (3x - 2) ]
dy/dx = 3 sin (6x - 4)

Let u = 3x - 2
du/dx
Let y = Sin^2(u) = Sin(u) Sin(u)
Apply the Product Rule which is
dy/dx = udv + v du
Hence
dy/du = Sin(u)Cos(u) + Cos(u) Sin(u)
dy/du = 2Sin(u)Cos(u)

Now apply the Chain Rule
dy/dx = dy/du X du/dx
dy/dx = 2Sin(u)Cos(u) X 2
dy/dx = 6Sin(u)Cos(u)
Substitute back in for 'u'
dy/dx = 6Sin(3x -2)Cos(3x - 2)

Using the Trig. Identity
3(2Sin(3x-2)(Cos(3x-2)) =
3(Sin(3x-2)Cos(3x-2) + Cos(3x-2)Sin(3x-2)) =>
3Sin((3x - 2) + (3x - 2)) =>
3Sin(6x - 4)
Hence
dy/dx = 3Sin(6x - 4)

Chain rule gives you the LHS. You really need to apply it 3 times.
d/dx f^2 = 2f giving you 2sin
Then d/dx sin is cos giving you 2sin*cos
Then d/dx 3x-2 is 3. Put that all together and you get
2sin(3x-2)cos(3x-2)3
or
6sin(3x-2)cos(3x-2)

Double angle formula gives you the rest.