Step 1: Factor out the variables.
Take $x$ from row 1, $y$ from row 2 and $z$ from row 3. The determinant becomes $xyz$ times a new determinant.
\[ xyz\begin{vmatrix} 1+\frac1x & \frac1x & \frac1x \\ \frac1y & 1+\frac1y & \frac1y \\ \frac1z & \frac1z & 1+\frac1z \end{vmatrix} \]
Step 2: Add the columns.
Add columns 2 and 3 to column 1 ($C_1 \to C_1+C_2+C_3$). Every entry of the new first column equals $1+\frac1x+\frac1y+\frac1z$.
Step 3: Take the common factor out.
Let $S = 1+\frac1x+\frac1y+\frac1z$. Then the determinant is $xyz \cdot S \cdot D'$, where $D'$ has first column all 1s.
Step 4: Evaluate D'.
Subtract row 1 from rows 2 and 3 in $D'$. This makes the first column $(1,0,0)^T$ and leaves the lower-right block as the identity. So $D' = 1$.
Step 5: Solve.
The determinant is $xyz\cdot S = 0$. As $xyz \neq 0$, we get $S=0$.
\[ \frac1x+\frac1y+\frac1z = -1 \]
Final Answer:
The required sum is $-1$.
\[ \boxed{-1} \]