En primer lugar, eliminamos los paréntesis. Vamos de afuera hacia adentro:
\[
\begin{aligned}
\frac{1}{a} – \left( \frac{1}{b} – \left( \frac{a}{b} – \frac{b}{a} \right) \right)
&=
\frac{1}{a} – \frac{1}{b} + \left( \frac{a}{b} – \frac{b}{a} \right)
\\[1em]
&=
\frac{1}{a} – \frac{1}{b} + \frac{a}{b} – \frac{b}{a}
\end{aligned}
\]
\[
\begin{aligned}
&\frac{1}{a} – \left( \frac{1}{b} – \left( \frac{a}{b} – \frac{b}{a} \right) \right)
\\[1em]
&\hspace{6em}=
\frac{1}{a} – \frac{1}{b} + \left( \frac{a}{b} – \frac{b}{a} \right)
\\[1em]
&\hspace{6em}=
\frac{1}{a} – \frac{1}{b} + \frac{a}{b} – \frac{b}{a}
\end{aligned}
\]
Ahora reemplazamos los valores \( a = 2 \) y \( b = 3 \):
\[
\begin{aligned}
\frac{1}{a} – \left( \frac{1}{b} – \left( \frac{a}{b} – \frac{b}{a} \right) \right)
&=
\frac{1}{a} – \frac{1}{b} + \frac{a}{b} – \frac{b}{a}
\\[1em]
&=
\frac{1}{2} – \frac{1}{3} + \frac{2}{3} – \frac{3}{2}
\\[1em]
&=
-\frac{2}{2} + \frac{1}{3}
\\[1em]
&=
-\frac{ -6 + 2 }{6}
\\[1em]
&=
\frac{-4}{6}
\\[1em]
&= \boldsymbol{ -\frac{2}{3} }
\end{aligned}
\]
\[
\begin{aligned}
&\frac{1}{a} – \left( \frac{1}{b} – \left( \frac{a}{b} – \frac{b}{a} \right) \right)
\\[1em]
&\hspace{5em}=
\frac{1}{a} – \frac{1}{b} + \frac{a}{b} – \frac{b}{a}
\\[1em]
&\hspace{5em}=
\frac{1}{2} – \frac{1}{3} + \frac{2}{3} – \frac{3}{2}
\\[1em]
&\hspace{5em}=
-\frac{2}{2} + \frac{1}{3}
\\[1em]
&\hspace{5em}=
-\frac{ -6 + 2 }{6}
\\[1em]
&\hspace{5em}=
\frac{-4}{6}
\\[1em]
&\hspace{5em}= \boldsymbol{ -\frac{2}{3} }
\end{aligned}
\]
