I recommend you go back and check the problem. I suspect it should be \(\displaystyle x^2- 6x+ y^2+ 4y= 3\).
If \(\displaystyle x^2- 6x+ y^2+ 4x= 3\) really is the problem, -6x+ 4x= -2x so this is the same as \(\displaystyle x^2- 2x+ y^2= 3\). The fact that there is no "y to the first power" term is not a problem. It only means that y is already a "perfect square""\(\displaystyle y^2= (y- 0)^2\). Complete the square in x.