in order to find the equation of the tangent line you need to slope which is equal to the derivative. use implicit differentiation
2x + y + x(dy/dx) - 3y^2(dy/dx) = y^2 + x(2y)(dy/dx)
solve for dy/dx
x(dy/dx) - 3y^2(dy/dx) - 2xy(dy/dx) = -2x - y + y^2
(dy/dx)(x - 3y^2 - 2xy) = -2x - y + y^2
then dy/dx = [-2x - y + y^2]/ [x - 3y^2 - 2xy]
now to find the value of the slope use your point (1,1) for x and y
dy/dx = [-2(1) - 1 + 1^2]/ [1 - 3(1^2) - 2(1)(1)]
dy/dx = -2/-4 so dy/dx = 1/2
now use point-slope with m=1/2 and the point (1,1)
y - 1 = 1/2(x -1)
y - 1 = 1/2x - 1/2
y = 1/2x + 1/2 this is the equation of your tangent line.