A LEVEL MATHS help

I found a question in the textbook that I'm struggling to answer.
Simplify this expression by expanding brackets, factorising or both:
[(2x+y)(y-2x)]²-(4x²+y²+2z²)²
At the back of the book with the answer it gives the advice that it is the difference of two squares, so I attempted to do it like that but I got a different answer from the textbook.
The answer in the textbook is -4(4x²+z²)(y²+z²)
Can someone please help explain where I went wrong I will post my working out

So this is how I worked out my answer:
[((2x+y)(y-2x))-(4x²+y²+2z²)][((2x+y)(y-2x))+(4x²+y²+2z²)]
[(y²-4x²)-(4x²+y²+2z²)(y²-4x²)+(4x²+y²+2z²)]
(-8x²-2z²)(2y²+2z²)
-2(4x²+z²)(-y²-z²)

So this is how I worked out my answer:
[((2x+y)(y-2x))-(4x²+y²+2z²)][((2x+y)(y-2x))+(4x²+y²+2z²)]
[(y²-4x²)-(4x²+y²+2z²)(y²-4x²)+(4x²+y²+2z²)]
(-8x²-2z²)(2y²+2z²)
-2(4x²+z²)(-y²-z²)

Youre not miles off, but Id have done dots on
((2x+y)(y-2x))
first, then dots again, which youve sort of done. But at the end you have both a 2 and a -2 factor which gives -4 in front of the remaining expression.

Oh so it's
(-8x²-2z²)(2y²+2z²)
-2(4x²+z²)2(y²+z²)
Then -2x2 is -4 so -4(4x²+z²)(y²+z²)?

Yes. Algebraically if you had
2ab
you could interpret that as either
(2a)b or a(2b)
so if you have (2a)(2b) then its 4ab