$\mathit{f}\left(\mathit{x}\right)=\frac{4\mathit{x}+3}{6\mathit{x}-4}$
$\mathit{f}\circ \mathit{f}\left(\mathit{x}\right)=\frac{4\left(\frac{4\mathit{x}+3}{6\mathit{x}-4}\right)+3}{6\left(\frac{4\mathit{x}+3}{6\mathit{x}-4}\right)-4}$
$=\frac{16\mathit{x}+12+18\mathit{x}-12}{24\mathit{x}+18-24+16}$
$=\frac{34\mathit{x}}{34}=\mathit{x}$
$\mathit{y}=\frac{4\mathit{x}+3}{6\mathit{x}-4}$
$6\mathit{x}\mathit{y}-4\mathit{y}=4\mathit{x}+3$
$6\mathit{x}\mathit{y}-4\mathit{x}=4\mathit{y}+3$
$\mathit{x}=\frac{4\mathit{y}+3}{6\mathit{y}-4}$
$\therefore \mathit{f}\left(\mathit{x}\right)=\frac{4\mathit{x}+3}{6\mathit{x}-4}$
OR
Given :
Let $\mathit{A}$ be the set of real numbers.
(i) For symmetric : Let $\mathit{a},\mathit{b}\in \mathrm{A}$
Then, $\left(\mathit{a},\mathit{b}\right)\in \mathrm{R}$ such that
which is false
$\Rightarrow \left(\mathit{b},\mathit{a}\right)\notin \mathrm{R}$
$\therefore \mathrm{R}$ is not symmetric.
(ii) For transitive : Let $\mathit{a},\mathit{b},\mathit{c}\in \mathrm{A}$
Then, $\left(\mathit{a},\mathit{b}\right)\in \mathrm{R}$ and $\left(\mathit{b},\mathit{c}\right)\in \mathbf{R}$


$\Rightarrow \left(\mathit{a},\mathit{c}\right)\in \mathrm{R}$
$\therefore \mathrm{R}$ is transitive.