$(2x^2+y^2-5)(2xy-1)$ is positive or negative. Sketch a Picture showing the regions where  $(2x^2+y^2-5)(2xy-1)$ is positive or negative. Indicate where the expression is 0 and areas that are not defined if they exist.

I know this is not the ideal way to ask a question, but I am unsure about how the function is graphed as shown in the first graph of the image above. How do I even attempt this question? Any guidance is appreciated.
 A: Considering that
$$
e(x,y)=2x^2+y^2=5\\
h(x,y)=2 x y = 1
$$
represent respectively an ellipse and an equilateral hyperbola, we can easily establish the positive and negative regions. So we have
$$
e(x,y) < 0 \to\ \ \ \mbox{ellipse internal points }\\
e(x,y) > 0 \to\ \ \ \mbox{ellipse external points}\\
$$ 
etc.
First $e(x,y)h(x,y) > 0$ (light blue)

and following, $e(x,y)h(x,y) < 0$ (light blue)

and in black $e(x,y)h(x,y) = 0$
A: Let us set $f(x,y) = (2x^2+y^2+5)(2xy-1)$. Then you want to sketch 
(i) the part of the $x-y$ plane s.t. $f(x,y)$ is positive
(ii) the part of the $x-y$ plane s.t. $f(x,y)$ is negative
(iii) the part of the $x-y$ plane s.t. $f(x,y)$ is 0.
As far as doing this HINT: Graph the curves $2x^2 +y^2 = 5$ (it is an elipse centred at the origin) and $2xy-1 = 0$ (there is a curve in each of the two quadrants $x >0,y>0$ and $x<0,y<0$; in fact the graph for $2xy-1 =0$ is the same as the graph for $y=\frac{1}{2x}$). 
The curves  $2x^2 +y^2 = 5$ and $xy-1 =0$ form the boundary between the regions $\{(x,y): f(x,y)>0\}$ and $\{(x,y): f(x,y) <0\}$ [can you figure which is which]. And in fact $\{(x,y): f(x,y)=0\}$ is precisely the set of points $(x,y)$ satisfying either $2x^2 +y^2 = 5$ or $2xy-1 = 0$ i.e., a point on one of the curves that you just graphed. Can you see why and how.
A: In general if you are asked to draw $R(x,y)=0$ and you can solve so that $R(x,y)=0$ if and only if $y=f(x)$. Then you draw the graph for $f$ as usual. Everything on the lines of the graph satisfies $R(x,y)=0$.
Here you want $R(x,y)=(2x^2+y^2-5)(2xy-1)$. This is true when either $(2x^2+y^2-5=0)$ or $2xy-1=0$. 
Solve for $f$ for the second equation. $y = \frac{1}{2x}$. So draw that on your graph paper.
Now for the other equation. We can't solve for a function $y=f(x)$ anymore. $y = \pm \sqrt{5-2x^2}$ provided $2x^2 \leq 5$. The $\pm$ and $2x^2 \leq 5$ conditions make it not a function but you can still draw both sides of this ellipse on your same graph paper.
Now you have all the places where $R(x,y)=0$ are marked. This breaks up the plane into regions. Each region will have some sign. The changes of signs can only occur at the graphs you've drawn as the value might cross through $0$.
So pick a point in each of the region to figure out that the sign is there.
