The value of the expression 

The correct option is 1.

By analyzing the provided images, we identify key logarithmic properties that allow us to simplify the expression:
1. Identity property: ${\log }_a\left(a\right)=1$
2. Product rule: ${\log }_a\left(xy\right)={\log }_a\left(x\right)+{\log }_a\left(y\right)$
3. Reciprocal/change of base property: ${\log }_a\left(b\right)=\frac{1}{{\log }_b\left(a\right)}$

Let us simplify each denominator in the expression one by one. Starting with the first term:
$$1+{\log }_u\left(vw\right)={\log }_u\left(u\right)+{\log }_u\left(vw\right)$$
Using the product rule, we combine the logarithm terms:
$${\log }_u\left(u\right)+{\log }_u\left(vw\right)={\log }_u\left(uvw\right)$$

Similarly, we simplify the second and third denominators:
$$1+{\log }_v\left(wu\right)={\log }_v\left(v\right)+{\log }_v\left(wu\right)={\log }_v\left(uvw\right)$$
and
$$1+{\log }_w\left(uv\right)={\log }_w\left(wu\right)+{\log }_w\left(uv\right)={\log }_w\left(uvw\right)$$

Now, substituting these simplified denominators back into the main expression gives:
$$\frac{1}{{\log }_u\left(uvw\right)}+\frac{\mathrm{1<!--\; <mrow-->\; <msub>\; <mi>log</mi>\; <mi>v</mi>\; </msub>\; <mo>(</mo>\; <mi>u</mi>\; <mi>v</mi>\; <mi>w</mi>\; <mo>)</mo>}}{}+\frac{1}{{\log }_w\left(uvw\right)}$$

Using the reciprocal logarithmic property, $\frac{1}{{\log }_a\left(b\right)}={\log }_b\left(a\right)$, we can rewrite each term as follows:
$${\log }_{uvw}\left(u\right)+{\log }_{uvw}\left(v\right)+{\log }_{uvw}\left(w\right)$$

Applying the product rule of logarithms to combine these terms under the common base $uvw$:
$${\log }_{uvw}(u\cdot v\cdot w)={\log }_{uvw}\left(uvw\right)$$

Since the argument is identical to the base, the logarithm evaluates to 1:
$${\log }_{uvw}\left(uvw\right)=1$$