If $E/ \mathbb{Q}$ elliptic curve in the general Form of Weierstrass and $P=(x,y)$ a rational point of it, show that the first coordinate of the point $2P$ is
$$ x(2P)=\frac{x^4-b_4x^2-2b_6x-b_8}{4x^3+b_2x^2+2b_4x+b_6}$$ where $$b_2=a_1^2+4a_2$$ $$b_4=a_1a_3+2a_4$$ $$b_6=a_3^2+4a_6$$ $$b_8=a_1^2a_6-a_1a_3a_4+4a_2a_6+a_2a_3^2-a_4^2$$
$$$$
I have done the following:
$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$, then
$$\lambda=\frac{3x_1^2+2a_2x_1+a_4-a_1y_1}{2y_1+a_1x_1+a_3}, v=\frac{-x_1^3+a_4x_1+2a_6-a_3y_1}{2y_1+a_1x_1+a_3}$$ and then $$2P=(\lambda^2+a_1 \lambda-a_2-x_1-x_2, -(\lambda+a_1)x_3-v-a_3)$$ $P_1=P_2 \Rightarrow x_1=x_2=x, y_1=y_2=y$ $$\lambda^2+a_1 \lambda-a_2-x-x= \\ \left ( \frac{3x^2+2a_2x+a_4-a_1y}{2y+a_1x+a_3} \right )^2+a_1 \frac{3x^2+2a_2x+a_4-a_1y}{2y+a_1x+a_3}-a_2-2x \\ =\frac{(3x^2+2a_2x+a_4-a_1y)^2}{(2y+a_1x+a_3)^2} +a_1 \frac{(3x^2+2a_2x+a_4-a_1y)(2y+a_1x+a_3)}{(2y+a_1x+a_3)^2}-\frac{(a_2+2x)(2y+a_1x+a_3)^2}{(2y+a_1x+a_3)^2} \\ = \frac{12a_2x^3-6a_1x^2y+4a_2^2x^2+6a_4x^2-4a_1a_2xy+4a_2a_4x+a_1^2y^2-2a_1a_4y+a_4^2+9x^4}{(2y+a_1x+a_3)^2}+\frac{3a_1^2x^3+6a_1x^2y+2a_1^2a_2x^2+3a_1a_3x^2-a_1^3xy+4a_1a_2xy+2a_1a_2a_3x+a_1^2a_4x-2a_1^2y^2-a_1^2a_3y+2a_1a_4y+a_1a_2a_3}{(2y+a_1x+a_3)^2}-\frac{2a_1^2x^3+8a_1x^2y+a_1^2a_2x^2+4a_1a_3x^2+4a_1a_2xy+8a_3xy+2a_3^2x+2a_1a_2a_3x+4a_2y^2+4a_2a_3y+a_2a_3^2+8xy^2}{(2y+a_1x+a_3)^2}$$
Is it right so far? How can we continue in order to show the desired result?
EDIT:
EDIT 2
How can we find the tangent of the curve at the point $P$ ?
In the first line you already have given the general Weierstrass form for the affine curve. The other one is just the associated projective curve, where you obtain the affine part by setting $z=1$. So the difference is only whether you want to consider the affine or projective Weierstrass form for your elliptic curve. Usually, the affine form is used, with the formula for $2P$, as you have written (with $P=(x_1,y_1)$ ?). Note that there are typos in your formula, e.g. $2_3y_1$, and what are $x_2,x_3$ ?