One-one/many-one : Let $\mathrm{x}_1, \mathrm{x}_2 \in \mathrm{R}-\{3\}$ are the elements such that
$f\left(x_1\right)=f\left(x_2\right)$ : then $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow \frac{\mathrm{x}_1-2}{\mathrm{x}_1-3}=\frac{\mathrm{x}_2-2}{\mathrm{x}_2-3}$ $\Rightarrow\left(\mathrm{x}_1-2\right)\left(\mathrm{x}_2-3\right)=\left(\mathrm{x}_2-2\right)\left(\mathrm{x}_1-3\right)$ $\Rightarrow \mathrm{x}_1 \mathrm{x}_2-2 \mathrm{x}_2-3 \mathrm{x}_1+6=\mathrm{x}_2 \mathrm{x}_1-2 \mathrm{x}_1-3 \mathrm{x}_2+6$ $\Rightarrow-2 \mathrm{x}_2-3 \mathrm{x}_1=-2 \mathrm{x}_1-3 \mathrm{x}_2$ $\Rightarrow \mathrm{x}_2=\mathrm{x}_1, \therefore \mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \mathrm{x}_1=\mathrm{x}_2$ $\Rightarrow \mathrm{f}$ is one-one function
Onto/Into : Let $\mathrm{y} \in \mathrm{R}-\{1\}$ (co-domain)
Then one element $\mathrm{x} \in \mathrm{R}-\{3\}$ in domain is such that
$$
f(x)=y \Rightarrow \frac{x-2}{x-3}=y \Rightarrow x-2=x y-3 y
$$
$$
\Rightarrow x=\left(\frac{3 y-2}{y-1}\right)
$$
$\therefore$ The pre-image of each element of co-domain $\mathrm{R}-\{1\}$ exists in domain $\mathrm{R}-\{3\}$.
$\Rightarrow \mathrm{f}$ is onto