To determine the total number of second-order partial derivatives for a function of three variables $f(x, y, z)$, we use the fact that we can take derivatives with respect to each variable twice. In general, for a function of $n$ variables, the number of second-order partial derivatives is given by the combination formula for choosing 2 items from $n$ (with repetition), which is:

$\binom{n+2-1}{2} = \binom{n+1}{2} = \frac{(n+1)n}{2}$

For $n = 3$ (since we have three variables: $x, y, z$):

$\binom{3+1}{2} = \binom{4}{2} = \frac{4 \cdot 3}{2} = 6$

Thus, there are 6 second-order partial derivatives. These are:

1. $\frac{\partial^2 f}{\partial x^2}$
2. $\frac{\partial^2 f}{\partial y^2}$
3. $\frac{\partial^2 f}{\partial z^2}$
4. $\frac{\partial^2 f}{\partial x \partial y}$ (or $\frac{\partial^2 f}{\partial y \partial x}$)
5. $\frac{\partial^2 f}{\partial x \partial z}$ (or $\frac{\partial^2 f}{\partial z \partial x}$)
6. $\frac{\partial^2 f}{\partial y \partial z}$ (or $\frac{\partial^2 f}{\partial z \partial y}$)

Since mixed partial derivatives are equal if the function and its partials are continuous (Schwarz's theorem or Clairaut's theorem), the number of distinct second-order partial derivatives is 6.

Next, we calculate the mixed partial derivative $\frac{\partial^4 f}{\partial z^2 \partial x \partial y}$.

Given $f(x, y, z) = 3x^2yz^4 + 2xz \cos(xy)$:

1. First partial derivative with respect to $y$:

$f_y = \frac{\partial}{\partial y} (3x^2yz^4 + 2xz \cos(xy)) = 3x^2z^4 + 2xz \cdot (-x \sin(xy)) = 3x^2z^4 - 2x^2z \sin(xy)$

2. Second partial derivative with respect to $x$:

$\frac{\partial}{\partial x} \left(3x^2z^4 - 2x^2z \sin(xy)\right) = 6xz^4 - 2z \sin(xy) \cdot 2x - 2x^2z \cos(xy) \cdot y$ $= 6xz^4 - 4xz \sin(xy) - 2x^2z y \cos(xy)$

3. Third partial derivative with respect to $z$:

$\frac{\partial}{\partial z} \left(6xz^4 - 4xz \sin(xy) - 2x^2z y \cos(xy)\right) = 6x \cdot 4z^3 - 4x \sin(xy) - 2x^2 y \cos(xy)$ $= 24xz^3 - 4x \sin(xy) - 2x^2 y \cos(xy)$

4. Fourth partial derivative with respect to $z$:

$\frac{\partial}{\partial z} \left(24xz^3 - 4x \sin(xy) - 2x^2 y \cos(xy)\right) = 24x \cdot 3z^2 - 0 - 0$ $= 72xz^2$

Thus, the fourth partial derivative $\frac{\partial^4 f}{\partial z^2 \partial x \partial y}$ is:

$\frac{\partial^4 f}{\partial z^2 \partial x \partial y} = 72xz^2$

Would you like more details or have any questions about this problem? Here are five questions you might want to ask next:

1. How do you find the first-order partial derivatives of $f(x, y, z)$?
2. What is Schwarz's theorem and how is it applied?
3. Can you explain the combination formula used to determine the number of second-order partial derivatives?
4. How do higher-order partial derivatives differ in their calculation?
5. Can you demonstrate a different example involving mixed partial derivatives?

Tip: When dealing with partial derivatives, ensure to follow the order of differentiation correctly and remember the rules of differentiation apply similarly to multivariable functions as they do in single-variable calculus.