To solve the problem, we need to find all ordered triples $(x, y, z)$ of positive integers such that $(x^y)^z = 64$. We start by expressing 64 as a power of a prime number:

$$64 = 2^6$$

This means we need to find all combinations of $x$, $y$, and $z$ such that:

$$(x^y)^z = 2^6$$

This can be rewritten as:

$$x^{yz} = 2^6$$

Since $x$ must be a power of 2 (because 64 is a power of 2), let $x = 2^a$ where $a$ is a positive integer. Substituting $x = 2^a$ into the equation, we get:

$$(2^a)^{yz} = 2^6$$

This simplifies to:

$$2^{ayz} = 2^6$$

Therefore, we must have:

$$ayz = 6$$

Now, we need to find all positive integer solutions $(a, y, z)$ such that $ayz = 6$. We will consider each possible value of $a$ and find the corresponding pairs $(y, z)$.

1. **For $a = 1$:**
$$yz = 6$$
The possible pairs $(y, z)$ are:
$$(1, 6), (2, 3), (3, 2), (6, 1)$$
This gives us 4 solutions.

2. **For $a = 2$:**
$$yz = 3$$
The possible pairs $(y, z)$ are:
$$(1, 3), (3, 1)$$
This gives us 2 solutions.

3. **For $a = 3$:**
$$yz = 2$$
The possible pairs $(y, z)$ are:
$$(1, 2), (2, 1)$$
This gives us 2 solutions.

4. **For $a = 6$:**
$$yz = 1$$
The possible pair $(y, z)$ is:
$$(1, 1)$$
This gives us 1 solution.

Summarizing all the solutions, we have:

$$4 + 2 + 2 + 1 = 9$$

Thus, the total number of ordered triples $(x, y, z)$ that satisfy the equation $(x^y)^z = 64$ is:

$\boxed{9}$