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A function $f: A \rightarrow B$ is onto (surjective) if for every element $y$ in the codomain $B$, there exists at least one element $x$ in the domain $A$ such that $f(x) = y$.
Let $y$ be an arbitrary element in the codomain $R$. We set $f(x) = y$:
$$y = \frac{x-2}{x-3}$$Multiply both sides by $(x-3)$:
$$y(x-3) = x-2$$ $$xy - 3y = x - 2$$ $$xy - x = 3y - 2$$ $$x(y-1) = 3y - 2$$ $$x = \frac{3y-2}{y-1}$$For $f$ to be onto, $x$ must exist in the domain $R - \{3\}$ for every $y \in R$. Looking at the expression $x = \frac{3y-2}{y-1}$, we see that $x$ is undefined when $y = 1$. This means there is no $x$ in the domain such that $f(x) = 1$.
Since $y=1$ is in the codomain $R$ but has no pre-image in the domain, the function is not onto.
Final Answer: The function is not onto.