$\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - 3{y^2} + 2xy - 2x - 10y + 4 = 0\\
{x^2} + {y^2} = 10
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 3{y^2} + 2xy - 2x - 10y + 4 = 0\left( 1 \right)\\
{x^2} = 10 - {y^2}\left( 2 \right)
\end{array} \right.\\
\left( 2 \right) \to \left( 1 \right):\left( {10 - {y^2}} \right) - 3{y^2} + 2xy - 2x - 10y + 4 = 0\\
 \Leftrightarrow 10 - 4{y^2} + 2xy - 2x - 10y + 4 = 0\\
 \Leftrightarrow 10 - 10y + 2x\left( {y - 1} \right) - 4{y^2} + 4 = 0\\
 \Leftrightarrow 10\left( {1 - y} \right) + 2x\left( {y - 1} \right) + 4\left( {1 - {y^2}} \right) = 0\\
 \Leftrightarrow 10\left( {1 - y} \right) + 2x\left( {y - 1} \right) - 4\left( {{y^2} - 1} \right) = 0\\
 \Leftrightarrow  - 4\left( {y - 1} \right)\left( {y + 1} \right) + 2x\left( {y - 1} \right) - 10\left( {y - 1} \right) = 0\\
 \Leftrightarrow \left( {y - 1} \right)\left[ { - 4\left( {y + 1} \right) + 2x - 10} \right] = 0\\
 \Leftrightarrow \left( {y - 1} \right)\left( {2x - 4y - 14} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
y = 1\\
x - 2y - 7 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
y = 1\left( 3 \right)\\
x = 2y + 7\left( 4 \right)
\end{array} \right.\\
\left( 3 \right) \to \left( 2 \right):{x^2} = 10 - 1 = 9 \Rightarrow {x^2} = 9 \Rightarrow x =  \pm 3\\
\left( 4 \right) \to \left( 2 \right):{\left( {2y + 7} \right)^2} = 10 - {y^2}\\
 \Leftrightarrow 4{y^2} + 28y + 49 = 10 - {y^2}\\
 \Leftrightarrow 5{y^2} + 28y + 39 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
y =  - 3 \Rightarrow x = 1\\
y =  - \dfrac{{13}}{5} \Rightarrow x = \dfrac{9}{5}
\end{array} \right.\\
 \Rightarrow \left( {x;y} \right) = \left( {3;1} \right),\left( { - 3;1} \right),\left( {1; - 3} \right),\left( {\dfrac{9}{5}; - \dfrac{{13}}{5}} \right)
\end{array}$