Answer
a) $\frac{dy}{dx}=\frac{6x^2-y-1}{x}$
b) $\frac{dy}{dx}=4x-\frac{2}{x^2}$
c) Answer is the demonstration in "work step by step".
Work Step by Step
$ x + xy − 2x^3 = 2 $
(a)Taking derivative with respect to x
$\frac{d(x+xy-2x^3)}{dx} =\frac{d(2)}{dx}$
$\frac{dx}{dx}+\frac{d(xy)}{dx}- \frac{d(2x^3)}{dx}=0$
$1+x\frac{dy}{dx}+y \frac{dx}{dx}-2\frac{d(x^3)}{dx}=0$
$1+x\frac{dy}{dx}+y-2(3x^2)=0$
$1+x\frac{dy}{dx}+y-6x^2=0$
$x\frac{dy}{dx}=6x^2-y-1$
$\frac{dy}{dx}=\frac{6x^2-y-1}{x}$..................... eq(1)
(b). $x + xy − 2x^3 = 2$
$ xy=2+2x^3-x$
$y=\frac{2+2x^3-x}{x}=2x^{-1}+2x^2-1$ ............................ eq (2)
Taking derivative with respect to x
$\frac{dy}{dx}= \frac{d(2x^{-1})}{dx}+ \frac{d(2x^2)}{dx}-\frac{d(1)}{dx}$
$\frac{dy}{dx}= 2\frac{d(x^{-1})}{dx}+ 2\frac{d(x^2)}{dx}-0$
$\frac{dy}{dx}=-2x^{-2}+ 2(2x)=-\frac{2}{x^2}+4x=4x-\frac{2}{x^2}$
(c)Putting equation (2) in equation (1)
$\frac{dy}{dx}=\frac{6x^2-( \frac{2+2x^3-x}{x} )-1}{x}=\frac{6x^3-2-2x^3+x-x}{x^2}=\frac{4x^3-2}{x^2}=4x-\frac{2}{x^2}$
