Implicit differentiation for $x^2+y^2-x^2y^2=1$ while the slope of the curve must be ($0$) $x^2+y^2+cx^2y^2=1$ 
1- What happens to the curve when $c=-1$? Describe what appears on the screen. Can you prove it algebraically?
2- Find $y'$ by implicit differentiation. For the case $c=-1$, is your expression for $y'$ consistent with what you discovered in part (b)?
I've solved the first question and found that the curve is two parallel lines ($y=1$, $y=-1$); and proved that algebraically. My proof is as follows, please correct me if I'm wrong:
When $c=-1, x^2+y^2-x^2y^2=1$. Then, $y^2-x^2y^2=1-x^2$. Then, $y^2(1-x^2)=1-x^2$. Dividing both sides by $(1-x^2)$, $y^2=1, y=1 or -1$ when $x\ne1$
The problem is in the 2nd question. The derivative of that curve when $c=-1$ is $\left[ y'= \dfrac{y^2x-x}{y(1-x^2)}\right]$, but according to the solution of the first question the slope must be ($0$), how can we make the two solutions consistent with each other???

When $c=-1$ then the curve reduces to $(y^2-1)(x^2-1)=0.$ That means either $y= 1$ or $y=-1$ or $x=1$ or $x=-1.$ So the graph consists of two vertical lines and two horizontal lines.
When $y=1$ or $y=-1$ then the expression you've got for $y'$ becomes $0,$ as you'd expect given that those are two horizontal lines. When $x=1$ or $x=-1,$ then the denominator in the expression you've got for $y'$ becomes $0$ and the derivative is undefined, as you'd expect for two vertical lines.
