Quick warning that this is not the most elegant solution.
Let [tex]x=2019[/tex] and [tex]y=80^{1/5}[/tex], so the matrix whose determinant we want looks like this:
[tex]M = \begin{bmatrix} -\dfrac{x^2}y & \dfrac{(x+1)^2}{y^2} & -\dfrac{(x+2)^2}{y^3} \\\\ \dfrac{(x+3)^2}{y^4} & -\dfrac{(x+4)^2}{y^5} & \dfrac{(x+5)^2}{y^6} \\\\ -\dfrac{(x+6)^2}{y^7} & \dfrac{(x+7)^2}{y^8} & -\dfrac{(x+8)^2}{y^9} \end{bmatrix}[/tex]
Since it's a 3x3 matrix, we can just compute the determinant directly using a Laplace expansion. Along the first row, for instance, we get
[tex]\det(M) = -\dfrac{x^2}y \begin{vmatrix} -\dfrac{(x+4)^2}{y^5} & \dfrac{(x+5)^2}{y^6} \\\\ \dfrac{(x+7)^2}{y^8} & -\dfrac{(x+8)^2}{y^9} \end{vmatrix} \\\\ - \dfrac{(x+1)^2}{y^2} \begin{vmatrix} \dfrac{(x+3)^2}{y^4} & \dfrac{(x+5)^2}{y^6} \\\\ -\dfrac{(x+6)^2}{y^7} & -\dfrac{(x+8)^2}{y^9} \end{vmatrix} \\\\ - \dfrac{(x+2)^2}{y^3} \begin{vmatrix} \dfrac{(x+3)^2}{y^4} & -\dfrac{(x+4)^2}{y^5} \\\\ -\dfrac{(x+6)^2}{y^7} & \dfrac{(x+7)^2}{y^8} \end{vmatrix}[/tex]
[tex]\det(M) = -\dfrac{x^2}y \cdot \dfrac{(x+4)^2(x+8)^2-(x+5)^2(x+7)^2}{y^{14}} \\\\ - \dfrac{(x+1)^2}{y^2} \cdot \dfrac{(x+5)^2(x+6)^2 - (x+3)^2(x+8)^2}{y^{13}} \\\\ - \dfrac{(x+2)^2}{y^3} \cdot \dfrac{(x+3)^2(x+7)^2 - (x+4)^2(x+6)^2}{y^{12}}[/tex]
[tex]\det(M) = \dfrac{\text{numerator}}{y^{15}}[/tex]
where the numerator is
[tex]\bigg((x+5)^2(x+7)^2 - (x+4)^2(x+8)^2\bigg)x^2 \\ + \bigg((x+3)^2(x+8)^2 - (x+5)^2(x+6)^2\bigg) (x+1)^2 \\ + \bigg((x+4)^2(x+6)^2 - (x+3)^2(x+7)^2\bigg) (x+2)^2[/tex]
Now just simplify. The numerator reduces drastically to a constant, 216. Then
[tex]\det(M) = \dfrac{216}{\left(80^{1/5}\right)^{15}} = \dfrac{6^3}{80^3} = \left(\dfrac3{40}\right)^3 = \boxed{\dfrac{27}{64,000}}[/tex]
