The synthetic division table is:
$$ \begin{array}{c|rrr}3&3&-4&9\\& & 9& \color{black}{15} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{orangered}{24} \end{array} $$The solution is:
$$ \frac{ 3x^{2}-4x+9 }{ x-3 } = \color{blue}{3x+5} ~+~ \frac{ \color{red}{ 24 } }{ x-3 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{3}&3&-4&9\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}3&\color{orangered}{ 3 }&-4&9\\& & & \\ \hline &\color{orangered}{3}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&3&-4&9\\& & \color{blue}{9} & \\ \hline &\color{blue}{3}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 9 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrr}3&3&\color{orangered}{ -4 }&9\\& & \color{orangered}{9} & \\ \hline &3&\color{orangered}{5}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&3&-4&9\\& & 9& \color{blue}{15} \\ \hline &3&\color{blue}{5}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 15 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrr}3&3&-4&\color{orangered}{ 9 }\\& & 9& \color{orangered}{15} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{orangered}{24} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x+5 } $ with a remainder of $ \color{red}{ 24 } $.