How do I do this?

(3x2y3)(2xy2)2

------------------

      -2x2 y

$\large \frac{(3x\times2y^3)(2x y^2 )^2}{-2x^2y}$

$=\large \frac{(3x\times2y^3)(4x^2 y^4 )}{-2x^2y}$ 

$=-12xy^6$

  !

I think 'Laws of powers' would be clearer....

(3x2y3)(2xy2)2

------------------         =    3x^2y^3 (4x^2y^4) / -2x^2y  = 12x^4y^7/(-2x^2y) = -6x^2y^6

      -2x2 y