To find the doubling map of the elliptic curve $y^2=x^3+1$ I want to verify that the doubling map of the elliptic curve $y^2=x^3+1$ is given by
\begin{align}
 P=(x,y) &\to 2P \\
 (x,y) &\mapsto \left(\frac{x^4-8x}{4x^3+4}, \frac{2x^6+40x^3}{8y^3} \right)
\end{align}

For calculation flexibility we denote $P=(x',y')$. We will find the tangent line at $P$ and make intersect with the curve to find $2P$.
Differentiating the curve with respect to $x$, we get $$\frac{dy}{dx}=\frac{3x^2}{2y}.$$
So the tangent line at $P$ is given by $$y-y'=\frac{3x'^2}{2y'}(x-x') \Rightarrow y=\frac{3x'^2x-3x'^3+2y'^2}{2y'}.$$
Substituting $y$ in the curve, we get
\begin{align}
&\left(\frac{3x'^2x-3x'^3+2y'^2}{2y'} \right)^2=x^3+1 \\
&\Rightarrow (3x'^2x-3x'^3+2y'^2)^2=4y'^2x^3+4y'^2 \\
&\Rightarrow 9x'^4x^2+9x'^6+4y'^4+18x'^5x+12x'^3y'^2-12x'^2y'^2x=4y'^2x^3+4y'^2 \\
& \Rightarrow 4y'^2x^3-9x'^4x^2+(12x'^2y'^2-18x'^5)x+(4y'^2-9x'^6-4y'^4-12x'^3y'^2)=0, \cdots (1)
\end{align}
The elliptic passes through $(0,1)$, and hence plugging it in $(1)$, we get
$$4y'^2-9x'^6-4y'^4-12x'^3y'^2=0, \cdots (2)$$
But now I am out of way, how to solve for $x', y'$ ?

what is the apropriate way to solve my question ?


As hinted by @David Lui and @kelalaka, I would like to answer partially.
The elliptic curve is $y^2=x^3+1$. The tangent line at $P=(x',y')$ is $$L: y=\frac{3x'^2x-3x'^3+2y'^2}{2y'}.$$
Substituting $L$ in the elleiptic curve, we get
$$4y'^2x^3-9x'^4x^2+(12x'^2y'^2-18x'^5)x+(4y'^2-9x'^6-4y'^4-12x'^3y'^2)=0,$$
which is a cubic equation with doubale zero $x'$. If $b$ be another zero of this cubic equation, then by theory of equations we have the folowing relation:
\begin{align} &2x'+b=x'+x'+b=\frac{9x'^4}{4y'^2} \\ \Rightarrow &b=\frac{9x'^4}{4y'^2}-2x' \\ \Rightarrow &b=\frac{9x'^4}{4(x'^3+1)}-2x', ~~(\because y'^2=x'^3+1) \\
\Rightarrow &b=\frac{x'^4-8x'}{4x'^3+4}:=x   \end{align}
Now substituting $\frac{x'^4-8x'}{4x'^3+4}:=x $ in the tangent line and using $y'^2=x'^3+1$, we get
\begin{align}y=\frac{2x'^6+40x'^3}{8y'^3}.   \end{align}
Therefore $2P=\left(\frac{x^4-8x}{4x^3+4}, \frac{2x^6+40x^3}{8y^3} \right)$. This concludes the doubling map.
