The correct answer is $\boxed{0}$.
To solve this limit, we can use L’Hôpital’s rule. L’Hôpital’s rule states that if $\lim_{x\to a} \frac{f(x)}{g(x)}$ exists and $\lim_{x\to a} \frac{f'(x)}{g'(x)}$ also exists, then $\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}$.
In this case, we have
$$\begin{align}
\mathop {\lim }\limits_{\left( {{\text{x,}}\,{\text{y}}} \right) \to \left( {0,\,0} \right)} \frac{{{{\text{x}}^2} – {\text{xy}}}}{{\sqrt {\text{x}} – \sqrt {\text{y}} }} &= \mathop {\lim }\limits_{\left( {{\text{x,}}\,{\text{y}}} \right) \to \left( {0,\,0} \right)} \frac{{2\text{x}} – {\text{y}}}{{\frac{1}{{2\sqrt {\text{x}}}} – \frac{1}{{2\sqrt {\text{y}}}}}} \
&= \mathop {\lim }\limits_{\left( {{\text{x,}}\,{\text{y}}} \right) \to \left( {0,\,0} \right)} \frac{{2\text{x}} – {\text{y}}}{{\frac{1}{{2\sqrt {\text{x}}}} – \frac{1}{{2\sqrt {\text{y}}}}} \cdot \frac{{2\sqrt {\text{x}}}}{{2\sqrt {\text{x}}}} \
&= \mathop {\lim }\limits_{\left( {{\text{x,}}\,{\text{y}}} \right) \to \left( {0,\,0} \right)} \frac{{4\text{x}^2} – {\text{xy}}}{{\sqrt {\text{x}} – \sqrt {\text{y}}}} \
&= \mathop {\lim }\limits_{\left( {{\text{x,}}\,{\text{y}}} \right) \to \left( {0,\,0} \right)} \frac{{4\text{x}^2} – {\text{xy}} + {\text{xy}}}{{\sqrt {\text{x}} – \sqrt {\text{y}}} + {\sqrt {\text{x}} – \sqrt {\text{y}}}} \
&= \mathop {\lim }\limits_{\left( {{\text{x,}}\,{\text{y}}} \right) \to \left( {0,\,0} \right)} \frac{{4\text{x}^2} – {\text{xy}} + {\text{xy}} + {\text{xy}}}{{2\sqrt {\text{x}} – 2\sqrt {\text{y}}}} \
&= \mathop {\lim }\limits_{\left( {{\text{x,}}\,{\text{y}}} \right) \to \left( {0,\,0} \right)} \frac{{4\text{x}^2} + 2\text{xy}}}{{2\sqrt {\text{x}} – 2\sqrt {\text{y}}}} \
&= \mathop {\lim }\limits_{\left( {{\text{x,}}\,{\text{y}}} \right) \to \left( {0,\,0} \right)} \frac{{4\text{x}^2} + 2\text{xy}}{{2\sqrt {\text{x}}}} \cdot \frac{{2\sqrt {\text{x}}}}{{2\sqrt {\text{x}}}} \
&= \mathop {\lim }\limits_{\left( {{\text{x,}}\,{\text{y}}} \right) \to \left( {0,\,0} \right)} 2\text{x} + \text{y} \
&= 0 + 0 \
&= 0.
\end{align}$$
Therefore, the value of the limit is $\boxed{0}$.