To solve the expression

\[
\frac{x}{xy + 1} + x + 1 + \frac{y}{yz + 1} + y + 1 + \frac{z}{zx + 1} + z + 1,
\]

given that \(xyz = 1\), we'll start by simplifying each fraction.

1. **Substituting \(z\)**:
Since \(xyz = 1\), we can express \(z\) as \(z = \frac{1}{xy}\).

2. **Rewrite the terms**:

- For \(\frac{x}{xy + 1}\):
\[
xy + 1 = xy + 1 \quad (\text{no change})
\]
- For \(\frac{y}{yz + 1}\):
\[
yz + 1 = y\left(\frac{1}{xy}\right) + 1 = \frac{y}{xy} + 1 = \frac{y + xy}{xy}
\]
- For \(\frac{z}{zx + 1}\):
\[
zx + 1 = z\left(\frac{1}{y}\right) + 1 = \frac{1}{y} + 1 = \frac{1 + y}{y}
\]

3. **Substituting back into the equation**:
Putting this all together, we have:

\[
\frac{x}{xy + 1} + x + 1 + \frac{y + xy}{xy} + y + 1 + \frac{1 + y}{y}
\]

Now let's simplify further. Group terms appropriately and combine:

- The total constant is \(3\) (since we add \(x + 1 + y + 1 + z + 1\)).
- The fractions can be calculated individually:

\[
\frac{x}{xy + 1} + \frac{y}{yz + 1} + \frac{z}{zx + 1}
\]

4. **Final Result**:
Each of the fractions \(\frac{x}{xy + 1}\), \(\frac{y}{yz + 1}\), and \(\frac{z}{zx + 1}\) can be calculated explicitly using values of \(x, y, z\).

However, the key part is to see how these simplify given \(xyz = 1\).

Given extensive calculations, if you substitute specific values for \(x, y, z\) (like 1, 1, 1) or solve algebraically, you'd find a simplification.

Since the exact numbers or relationships can be numerous, the total can simplify significantly, often leading to simple fractions or integer results depending on specific values.

Please provide values or further context to narrow down specific outcome.