Example 7 Find all real number pairs $(x, y)$ that satisfy the equation $x^{2}+(y-1)^{2}+(x-y)^{2}=$ $\frac{1}{3}$. (Example 5 from [1])
Introducing undetermined parameters $u, v, w$, by the Cauchy-Schwarz inequality we have:
\[
\begin{array}{l}
\left(u^{2}+v^{2}+w^{2}\right)\left[x^{2}+(y-1)^{2}+(x-y)^{2}\right] \geqslant \\
{[u x+v(y-1)+w(x-y)]^{2} }=[(u+w) x+(v-
[v]^{2} }=[(u+w) x+(v-w) y-v]^{2}
\end{array}
\]

Let $u+w=0, v-w=0$, we can take $u=-1, v=w=1$, substituting into the above inequality gives:
\[
x^{2}+(y-1)^{2}+(x-y)^{2} \geqslant \frac{1}{3}
\]

Given that $x^{2}+(y-1)^{2}+(x-y)^{2}=\frac{1}{3}$, so $-x = y-1 = x-y$, solving gives $x=\frac{1}{3}, y=\frac{2}{3}$.