Two questions about curves How to know if the curve $x^6 + y^6 = x^4y$ is closed ? And how to know that the curve $y^2 = x^2\frac{1 - x}{ 1 + x}$ has a loop ? PS: When I ask this two questions, I'm not considering the usage of a computer program. Another consideration is that I used the "Multivariable-Calculus" tag for the questions but I think it's necessary more than that to answer them.
 A: In polar coordinates, $x = r \cos(\theta)$, $y = r \sin(\theta)$, the first curve becomes $$r^6 (\cos^6(\theta) + \sin^6(\theta))  = r^5 \cos^4(\theta) \sin(\theta)$$
One solution is $r = 0$ (the origin, an isolated point); otherwise 
$$ r = \dfrac{\cos^4(\theta) \sin(\theta)}{\cos^6(\theta) + \sin^6(\theta)}$$
Note that the denominator is bounded away from $0$, so this is a closed curve.
The second curve is symmetric about the $x$ axis:
$$ y = \pm x \sqrt{\frac{1-x}{1+x}}$$
for those $x$ such that $(1-x)/(1+x) \ge 0$, i.e. $-1 < x \le 1$.
We have $y=0$ at $x=0$ and $x=1$.  The loop consists of 
$y = +x \sqrt{(1-x)/(1+x)}$ in the upper half plane and 
$y = -x \sqrt{(1-x)/(1+x)}$ in the lower half plane, for $0 \le x \le 1$.
A: I'm trying to comment on the other answer but I am unable to. To answer our question in the comments, you only need to understand the terminology in the definition, you don't really need a full course in real analysis to understand what he is saying. Here is an answer that I think will help. (Feel free to delete this since it doesn't adequately answer the post. ) 
https://math.stackexchange.com/a/1340651
