Let's expand everything we've got and see what we're working with:
#y=-(2x-1)^2-x^2-2x+3#
expand #(2x-1)^2#
#y=-((2x-1) xx (2x-1)) -x^2-2x+3#
#y=-(4x^2-2x-2x+1) - x^2 -2x +3#
distribute the negative
#y=-4x^2+4x-1-x^2-2x+3#
combine like-terms
#y=-5x^2+2x+2#
Now, let's rewrite the standard form into vertex form. To do that, we need to complete the square
#y=-5x^2+2x+2#
factor out the negative #5#
#y=-5(x^2-2/5x-2/5)#
Now we take the middle term (#2/5#) and divide it by #2#. That gives us #1/5#. Now we square it, which gives us #1/25#. Now we have the value that will give us a perfect square. We add #1/25# to the equation but we cannot randomly introduce a new value in this equation! What we can do is add #1/25# and then subtract it #1/25#. That way, we haven't actually changed the value of the equation.
So, we have #y=-5(x^2-2/5x-2/5 +1/25-1/25)#
#y=-5(color(red)(x^2-2/5x+1/25) -2/5-1/25)#
rewrite as a perfect square
#y=-5((x-1/5)^2-2/5-1/25)#
combine constants
#y=-5((x-1/5)^2-11/25)#
multiply #-11/25# by #-5# to remove one of the parentheses
#y=-5(x-1/5)^2+11/5#
Now we've got the equation in vertex form.
From here, we can tell the vertex very easily:
#y=-5(xcolor(blue)(-1/5))^2+color(green)(11/5)#
Gives us #(-color(blue)(-1/5), color(green)(11/5))#, or #(1/5, 11/5)#
