The synthetic division table is:
$$ \begin{array}{c|rrrr}3&-3&10&-6&9\\& & -9& 3& \color{black}{-9} \\ \hline &\color{blue}{-3}&\color{blue}{1}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ -3x^{3}+10x^{2}-6x+9 }{ x-3 } = \color{blue}{-3x^{2}+x-3} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-3&10&-6&9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ -3 }&10&-6&9\\& & & & \\ \hline &\color{orangered}{-3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-3&10&-6&9\\& & \color{blue}{-9} & & \\ \hline &\color{blue}{-3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}3&-3&\color{orangered}{ 10 }&-6&9\\& & \color{orangered}{-9} & & \\ \hline &-3&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-3&10&-6&9\\& & -9& \color{blue}{3} & \\ \hline &-3&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 3 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}3&-3&10&\color{orangered}{ -6 }&9\\& & -9& \color{orangered}{3} & \\ \hline &-3&1&\color{orangered}{-3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-3&10&-6&9\\& & -9& 3& \color{blue}{-9} \\ \hline &-3&1&\color{blue}{-3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&-3&10&-6&\color{orangered}{ 9 }\\& & -9& 3& \color{orangered}{-9} \\ \hline &\color{blue}{-3}&\color{blue}{1}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -3x^{2}+x-3 } $ with a remainder of $ \color{red}{ 0 } $.