The correct answer is $\boxed{0}$.
To solve this, we can use the chain rule. The chain rule says that if $u$ is a function of $v$ and $v$ is a function of $x$, then $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial v} \cdot \frac{\partial v}{\partial x}$.
In this case, we have $u = \log \left( {\frac{{{{\text{x}}^2} + {{\text{y}}^2}}}{{{\text{x}} + {\text{y}}}}} \right)$, so $v = \frac{{{{\text{x}}^2} + {{\text{y}}^2}}}{{{\text{x}} + {\text{y}}}}}$ and $x$ is the independent variable.
Therefore, we can use the chain rule to find $\frac{\partial u}{\partial x}$ as follows:
\begin{align}
\frac{\partial u}{\partial x} &= \frac{\partial u}{\partial v} \cdot \frac{\partial v}{\partial x} \
&= \frac{2x}{{{{\text{x}}^2} + {{\text{y}}^2}} – \frac{2y}{{{x^2} + {y^2}}} \
&= \frac{2x – 2y}{{{{\text{x}}^2} + {{\text{y}}^2}}} \
&= \frac{2(x – y)}{{{{\text{x}}^2} + {{\text{y}}^2}}}
\end{align}
Similarly, we can find $\frac{\partial u}{\partial y}$ as follows:
\begin{align}
\frac{\partial u}{\partial y} &= \frac{\partial u}{\partial v} \cdot \frac{\partial v}{\partial y} \
&= \frac{2y}{{{{\text{x}}^2} + {{\text{y}}^2}} – \frac{2x}{{{x^2} + {y^2}}} \
&= \frac{2y – 2x}{{{{\text{x}}^2} + {{\text{y}}^2}}} \
&= \frac{2(y – x)}{{{{\text{x}}^2} + {{\text{y}}^2}}}
\end{align}
Therefore, the value of $\text{x} \cdot \frac{\partial u}{\partial x} + \text{y} \cdot \frac{\partial u}{\partial y}$ is:
\begin{align}
\text{x} \cdot \frac{\partial u}{\partial x} + \text{y} \cdot \frac{\partial u}{\partial y} &= \text{x} \cdot \frac{2(x – y)}{{{{\text{x}}^2} + {{\text{y}}^2}}} + \text{y} \cdot \frac{2(y – x)}{{{{\text{x}}^2} + {{\text{y}}^2}}} \
&= \frac{2(x^2 – xy + y^2)}{{{{\text{x}}^2} + {{\text{y}}^2}}} + \frac{2(xy – x^2 + y^2)}{{{{\text{x}}^2} + {{\text{y}}^2}}} \
&= \frac{2(x^2 + y^2 – x^2 – y^2)}{{{{\text{x}}^2} + {{\text{y}}^2}}} \
&= \frac{0}{{{{\text{x}}^2} + {{\text{y}}^2}}} \
&= 0
\end{align}