$\begin{array}{l} \left\{ \begin{array}{l} {x^3} - {y^3} + 3{x^2} + 6x - 3y + 4 = 0\left( 1 \right)\\ {x^2} + {y^2} - 3x = 1\left( 2 \right) \end{array} \right.\\ \left( 1 \right) \Leftrightarrow \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right) + \left( {{x^2} + xy + {y^2}} \right) + 2{x^2} - {y^2} - xy + 6x - 3y + 4 = 0\\ \Leftrightarrow \left( {x - y + 1} \right)\left( {{x^2} + xy + {y^2}} \right) + x\left( {x - y + 1} \right) + {x^2} - {y^2} + 5x - 3y + 4 = 0\\ \Leftrightarrow \left( {x - y + 1} \right)\left( {{x^2} + xy + {y^2}} \right) + x\left( {x - y + 1} \right) + \left( {x - y} \right)\left( {x + y} \right) + 5x - 3y + 4 = 0\\ \Leftrightarrow \left( {x - y + 1} \right)\left( {{x^2} + xy + {y^2}} \right) + x\left( {x - y + 1} \right) + \left( {x - y + 1} \right)\left( {x + y} \right) + 4\left( {x - y + 1} \right) = 0\\ \Leftrightarrow \left( {x - y + 1} \right)\left( {{x^2} + xy + {y^2} + x + x + y + 4} \right) = 0\\ \Leftrightarrow \left( {x - y + 1} \right)\left( {{x^2} + xy + {y^2} + 2x + y + 4} \right) = 0\\ \Leftrightarrow \left( {x - y + 1} \right)\left( {2{x^2} + 2xy + 2{y^2} + 2x + 2y + 4} \right) = 0\\ \Leftrightarrow \left( {x - y + 1} \right)\left[ {{{\left( {x + y} \right)}^2} + {{\left( {x + 1} \right)}^2+2} + {{\left( {y + 1} \right)}^2}} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l} y = x + 1\left( 3 \right)\\ \left[ {{{\left( {x + y} \right)}^2} + {{\left( {x + 1} \right)}^2} + {{\left( {y + 1} \right)}^2}}+2 \right] = 0(VL) \end{array} \right.\\ \left( 3 \right) \to \left( 2 \right):{x^2} + {\left( {x + 1} \right)^2} - 3x = 1\\ \Leftrightarrow 2{x^2} - x = 0\\ \Leftrightarrow x\left( {2x - 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0 \Rightarrow y = 1\\ x = \dfrac{1}{2} \Rightarrow y = \dfrac{3}{2} \end{array} \right.\\ \left( {x;y} \right) = \left( {0;1} \right),\left( {\dfrac{1}{2};\dfrac{3}{2}} \right) \end{array}$
