The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&1&-3&1&-3\\& & -1& 4& \color{black}{-5} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{5}&\color{orangered}{-8} \end{array} $$The solution is:
$$ \frac{ x^{3}-3x^{2}+x-3 }{ x+1 } = \color{blue}{x^{2}-4x+5} \color{red}{~-~} \frac{ \color{red}{ 8 } }{ x+1 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&-3&1&-3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 1 }&-3&1&-3\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 1 } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&-3&1&-3\\& & \color{blue}{-1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-1&1&\color{orangered}{ -3 }&1&-3\\& & \color{orangered}{-1} & & \\ \hline &1&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&-3&1&-3\\& & -1& \color{blue}{4} & \\ \hline &1&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 4 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-1&1&-3&\color{orangered}{ 1 }&-3\\& & -1& \color{orangered}{4} & \\ \hline &1&-4&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 5 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&-3&1&-3\\& & -1& 4& \color{blue}{-5} \\ \hline &1&-4&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}-1&1&-3&1&\color{orangered}{ -3 }\\& & -1& 4& \color{orangered}{-5} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{5}&\color{orangered}{-8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-4x+5 } $ with a remainder of $ \color{red}{ -8 } $.