Convexity of the graph of an implicit function Let $f(x,y)=0$ be an implicit function and suppose this defines a curve $\gamma$ on $\Bbb{R}^2$.
(If it is necessary, assume $f$ is a two variable polynomial.)


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*What are the conditions that guarantees that $\gamma$ is a simple closed curve?

*If it is given that the graph of this function is a simple closed curve, is there any condition that allows us to check the convexity of the region bounded by $\gamma$ ?
Specifically I am looking an answer for the second question but could not find any thing related to this.  
For example:  Is there any systematical way to determine the convexity of the simple closed curves like $y^4+y+2x^2=1$ and $y^2+2x^2y+x^2(x^2+1)=1$?
 A: Answer to the second part of my own question I'll use the following facts:

Boundary of a convex set is always a convex curve.
A simple closed regular plane curve is convex if and only if its curvature is either always non-negative or always non-positive.
Curvature of an implicit function $f(x,y)=0$ is given by $$\kappa=\dfrac{2f_{xy} f_x f_y-f_y^2f_{xx} -f_x^2 f_{yy}}{(f_x^2+f_y^2)^{3/2}}.$$

Hence we just need to look at the curvature of the curve $\gamma : f(x,y)=0.$
A: Some broad hints: 


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*Check that for each point $(x,y) \in \gamma$, there is a neighborhood around that point that is like (construct a homeomorphism) $B_{\delta}(\mbox{origin}) \cap A_x$ where $A_x$ is the x-axis.  If the curve intersects itself, for example, at the point of intersections, such a homeomorphism cannot exist.

*The bounded region is either $f>0$ or $f<0$.  Given your $f$, it would be easy to know which one is the unbounded one.  Once you know that, you can then check convexity of the bounded region D (let's say $D=\{(x,y) \in \mathbb{R}^2 |f<0\}$) in the usual way ($\forall v, w \in D$, $0<\lambda<1$, is $z=\lambda v + (1-\lambda)w \in D$?, i.e, is $f(z)<0$).  In fact, it would be sufficient to check whether $\forall v, w \in \gamma$ that $z \in D$.
