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)
