<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] There's a way to simplify the denominator without explicitly solving for $y$. Note that
\begin{align} 
\big(4y-(1+x)\big)^2 &= 16y^2 - 8(1+x)y + (1+x)^2 \\
&= 8\big(2y^2 - (1+x)y\big) + (1+x)^2 \\
&= -8x + (1+x)^2 \\
&= x^2 - 6x + 1 
\end{align}
Therefore
\begin{align} 
y' = \frac{y-1}{4y-(1+x)} &= \frac{(y-1)\big(4y-(1+x)\big)}{\big(4y-(1+x)\big)^2} \\
&= \frac{4y^2 - 4y - (1+x)y+(1+x)}{x^2-6x+1} \\
&= \frac{2\big(2y^2-(1+x)y\big)+(1+x)y - 4y+ (1+x)}{x^2-6x+1} \\
&= \frac{-2x+(1+x)y-4y+1+x}{x^2-6x+1} \\
&= \frac{1-x+(x-3)y}{x^2-6x+1}
\end{align}
</s>