<s>[INST] How has this result been obtained?? We have the quadratic equation
$$ 2y^2 -(1+x)y + x = 0 $$
The author derives the following result from the above equation -
$$ y' = \frac{y-1}{4y-1-x} = \frac{(x-3)y-x+1}{x^2-6x+1} $$
Now, I understand the first part. It has been obtained by simply differentiating w.r.t to $x$ and then solving for $y'$. But, I am unable to obtain the second part. 
I have tried lot of different substitutions, but haven't been able to arrive at the result. Please help!!
 [/INST] $$(x^2-6x+1)(y-1)-(4y-x-1)((x-3)y-x+1)\\
=-2x^2-4xy^2-4xy+6x+12y^2+2yx^2-6y\\
=(6-2x)(2y^2-(1+x)y+x)\\=0.$$
The identity is valid.

To establish it, solve the initial equation for $y$
$$y=\frac{x+1\pm\sqrt{x^2-6x+1}}4$$
and plug in the fraction, to get
$$y'=\frac{x-3\pm\sqrt{x^2-6x+1}}{\pm4\sqrt{x^2-6x+1}}=\frac{\pm(x-3)\sqrt{x^2-6x+1}+x^2-6x+1}{4(x^2-6x+1)}=\frac{4(x-3)y+x^2-6x+1-(x-3)(x+1)}{4(x^2-6x+1)}.$$
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