If $f:R - \left\{ {\frac{3}{5}} \right\} \to R$ be defined by $f(x) = \frac{{3x + 2}}{{5x - 3}}$, then

It is given that,, $f(x) = \frac{{3x + 2}}{{5x - 3}}$
Let $y = \frac{{3x + 2}}{{5x - 3}}$

$3x + 2 = 5xy - 3y \Rightarrow x(3 - 5y) = - 3y - 2$

$x = \frac{{3y + 2}}{{5y - 3}} \Rightarrow {f^{ - 1}}(x) = \frac{{3x + 2}}{{5x - 3}}$

$\therefore {f^{ - 1}}(x) = f(x)$