Answer
\[(a)\; y^{'}=\frac{-(4x+1+y)}{x}\]
\[(b)\;y^{'}=\frac{-(2x^2+1)}{x^2}\]
$\;\;\;\; \;\;\;\; \;\;\;\;\;\;(c)$ See below
Work Step by Step
$(a) \; 2x^2+x+xy=1$ ___(1)
Differentiate (1) implicitly with respect to $x$
$4x+1+xy^{'}+y=0$
$ y^{'}=\frac{-(4x+1+y)}{x}$ ___(2)
$(b)\;$ Solving (1) for $y$ in terms of $x$
$y=\frac{1-x-2x^2}{x}$ ___(3)
Differentiate (3) with respect to $x$ by quotient rule
$y^{'}=\Large\frac{[1-x-2x^2]^{'}x-[1-x-2x^2](x)^{'}}{x^2}$
$y^{'}=\Large\frac{(-1-4x)x-(1-x-2x^2)}{x^2}$
$y^{'}=\large\frac{-2x^2-1}{x^2}$
$y^{'}=\large\frac{-(2x^2+1)}{x^2}$
$(c)\;$ Substituting $y=\frac{1-x-2x^2}{x}$ from (3) in (2)
$ y^{'}=\Large\frac{-\left[4x+1+\large\left( \frac{1-x-2x^2}{x} \right)\right]}{x}$
$y^{'}=\Large\frac{-[4x^2+x+1-x-2x^2]}{x^2}$
$y^{'}=\large\frac{-(2x^2+1)}{x^2}$
Therefore, our solutions to parts $(a)$ and $(b)$ are consistent.