The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&-1&1&-4&0&3\\& & -1& 0& -4& \color{black}{-4} \\ \hline &\color{blue}{-1}&\color{blue}{0}&\color{blue}{-4}&\color{blue}{-4}&\color{orangered}{-1} \end{array} $$Because the remainder $ \left( \color{red}{ -1 } \right) $ is not zero, we conclude that the $ x-1 $ is not a factor of $ -x^{4}+x^{3}-4x^{2}+3$.
First we need to create a synthetic division table.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&-1&1&-4&0&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ -1 }&1&-4&0&3\\& & & & & \\ \hline &\color{orangered}{-1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&-1&1&-4&0&3\\& & \color{blue}{-1} & & & \\ \hline &\color{blue}{-1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}1&-1&\color{orangered}{ 1 }&-4&0&3\\& & \color{orangered}{-1} & & & \\ \hline &-1&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&-1&1&-4&0&3\\& & -1& \color{blue}{0} & & \\ \hline &-1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 0 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}1&-1&1&\color{orangered}{ -4 }&0&3\\& & -1& \color{orangered}{0} & & \\ \hline &-1&0&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&-1&1&-4&0&3\\& & -1& 0& \color{blue}{-4} & \\ \hline &-1&0&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}1&-1&1&-4&\color{orangered}{ 0 }&3\\& & -1& 0& \color{orangered}{-4} & \\ \hline &-1&0&-4&\color{orangered}{-4}& \end{array} $$Step 8 : 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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&-1&1&-4&0&3\\& & -1& 0& -4& \color{blue}{-4} \\ \hline &-1&0&-4&\color{blue}{-4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}1&-1&1&-4&0&\color{orangered}{ 3 }\\& & -1& 0& -4& \color{orangered}{-4} \\ \hline &\color{blue}{-1}&\color{blue}{0}&\color{blue}{-4}&\color{blue}{-4}&\color{orangered}{-1} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -1 }\right)$.