To find all second-order partial derivatives of the function $f(x, y, z) = \frac{x^2 y^6}{z^3} - 2x^6 z + 8y^{-3} x^4 + 4z^2$, we need to compute the following derivatives:

1. First, let's compute the first partial derivatives with respect to $x$, $y$, and $z$:

   $f_x = \frac{\partial}{\partial x} \left(\frac{x^2 y^6}{z^3} - 2x^6 z + 8y^{-3} x^4 + 4z^2\right)$

   $= \frac{2xy^6}{z^3} - 12x^5 z + 32y^{-3} x^3$

   $f_y = \frac{\partial}{\partial y} \left(\frac{x^2 y^6}{z^3} - 2x^6 z + 8y^{-3} x^4 + 4z^2\right)$

   $= \frac{6x^2 y^5}{z^3} - 24y^{-4} x^4$

   $f_z = \frac{\partial}{\partial z} \left(\frac{x^2 y^6}{z^3} - 2x^6 z + 8y^{-3} x^4 + 4z^2\right)$

   $= -\frac{3x^2 y^6}{z^4} - 2x^6 + 8z$

2. Now, compute the second-order partial derivatives:

   • Second partial derivatives with respect to $x$:

     $f_{xx} = \frac{\partial}{\partial x}(f_x) = \frac{\partial}{\partial x} \left( \frac{2xy^6}{z^3} - 12x^5 z + 32y^{-3} x^3\right)$

     $= \frac{2y^6}{z^3} - 60x^4 z + 96y^{-3} x^2$

   • Mixed partial derivatives with respect to $x$ and $y$:

     $f_{xy} = \frac{\partial}{\partial y}(f_x) = \frac{\partial}{\partial y} \left(\frac{2xy^6}{z^3} - 12x^5 z + 32y^{-3} x^3\right)$

     $= \frac{12xy^5}{z^3} - 96y^{-4} x^3$

   • Mixed partial derivatives with respect to $x$ and $z$:

     $f_{xz} = \frac{\partial}{\partial z}(f_x) = \frac{\partial}{\partial z} \left(\frac{2xy^6}{z^3} - 12x^5 z + 32y^{-3} x^3 \right)$

     $= -\frac{6xy^6}{z^4} - 12x^5$

   • Second partial derivatives with respect to $y$:

     $f_{yy} = \frac{\partial}{\partial y}(f_y) = \frac{\partial}{\partial y} \left(\frac{6x^2 y^5}{z^3} - 24y^{-4} x^4\right)$

     $= \frac{30x^2 y^4}{z^3} + 96y^{-5} x^4$

   • Mixed partial derivatives with respect to $y$ and $z$:

     $f_{yz} = \frac{\partial}{\partial z}(f_y) = \frac{\partial}{\partial z} \left(\frac{6x^2 y^5}{z^3} - 24y^{-4} x^4 \right)$

     $= -\frac{18x^2 y^5}{z^4}$

   • Second partial derivatives with respect to $z$:

     $f_{zz} = \frac{\partial}{\partial z}(f_z) = \frac{\partial}{\partial z} \left(-\frac{3x^2 y^6}{z^4} - 2x^6 + 8z\right)$

     $= \frac{12x^2 y^6}{z^5} + 8$

These are the second-order partial derivatives of the given function.