some functions, ploting them and defining positive and negative areas I have four following functions. I need to plot them and define positive and negative defined area. I have done most of them. But, i want you to check them, and correct them if false.

(1) $F(x,y)=x^2+2y^2$ 
for $x^2+2y^2=1$ , the graph is that 

$C=\{(x,y)| -1\lt x \lt 1\  and -1/\sqrt{2} \lt y \lt 1/\sqrt {2}\} $ 
i.e the set C refers to inside of the ellipse  
in this set C, the function is positive defined.

(2) $F(x,y)=xy$
for $xy=1$, the graph is that 

this is hyperbol. 
$C=\{(x,y)|  x \lt 0 \   and\  y\lt 0,\ x\gt 0 \  and \ y\gt 0 \} $ 
in this set C, the function is positive defined.

(3) $F(x,y)=x+2y^2$ 
for $x+2y^2=0$ the graph is that 

well, I cannot the set C which shows the positive and negative defined areas of the function.
please help me doing this.

(4) $F(x,y)=x^2+3xy+3y^2$
I can neither graph this function nor define the set C (positive and negatife defined areas).

Especailly, I cannot do part (3) and (4). 
Hopefully help me doing these two parts. And please check other two parts (1) and(2). Thank you so much.
 A: (4) 
The function $F(x,y)=x^2+3xy+3y^2$ can be thought as the quadratic form
$$F(x,y)=(x,y)\left(\begin{array}{cc}1&1\\2&3\end{array}\right)
\left(\begin{array}{c}x\\y\end{array}\right).\qquad (*)$$
Let us abbreviate $p=\left(\begin{array}{c}x\\y\end{array}\right)$ and
$Q=\left(\begin{array}{cc}1&1\\2&3\end{array}\right)$.
Now $Q$ isn't symmetric but can be decomposed as 
$$Q=\frac{1}{2}(Q+Q^{\top})+\frac{1}{2}(Q-Q^{\top}).$$ 
This decomposition gives us the matrices 
$$S=\frac{1}{2}(Q+Q^{\top})\quad \mbox{and}\quad A=\frac{1}{2}(Q-Q^{\top}),$$
which are $S$ symmetric and $A$ antisymmetric.
The matrix $S$ is 
$$\left(\begin{array}{cc}1&3/2\\3/2&3\end{array}\right).$$
Then ours expression  for $F$ is
$$F(p)=p^{\top}Qp=p^{\top}(S+A)p=p^{\top}Sp,\qquad (**)$$
because it is easy to compute $p^{\top}Ap=0$.
Also note that
$$F(x,y)=(x,y)\left(\begin{array}{cc}1&3/2\\3/2&3\end{array}\right)
\left(\begin{array}{c}x\\y\end{array}\right)=x^2+3xy+3y^2.$$
Then the sign of $F$ will depend on the symmetric quadratic form $p^{\top}Sp$.
To decide the sign of this function, knowing that the level curve $x^2+3xy+3y^2=1$ is a tilted ellipse, one should try to calculate a new basis for the vectorspace $\Bbb R^2$. 
This change gives a pair of axes that one gets from the eigenvectors of the matrix $S$.
The eigenvectors are 
$$v_1=-\frac{2+\sqrt{13}}{3}e_1+e_2,$$ 
and 
$$v_2=\frac{-2+\sqrt{13}}{2}e_1+e_2.$$
With these, one can take the matrix 
$$B=
\left(\begin{array}{cc}
-\frac{2+\sqrt{13}}{3}&-\frac{2-\sqrt{13}}{3}\\
1&1
\end{array}\right),$$
and to get the interpretation of the quadratic form in $(**)$ with the basis $\{v_1,v_2\}$, this will be through the formula $B^{\top}QB$ (this is dubbed orthogonal diagonalization for $Q$).
So the quadratic form for $S$ is
$$p^{\top}Sp=({BB^{-1}p})^{\top}SBB^{-1}p=(B^{-1}p)^{\top}(B^{\top}SB)B^{-1}p$$
Hence any can compute that the resulting expression is
\begin{eqnarray*}
(x,y)\left(\begin{array}{cc}1&3/2\\3/2&3\end{array}\right)
\left(\begin{array}{c}x\\y\end{array}\right)
&=&
(s,t)B^{\top}SB
\left(\begin{array}{c}s\\t\end{array}\right)\\
&=&
(s,t)
\left(\begin{array}{cc}
\frac{26-5\sqrt{13}}{9}&0\\
0&\frac{26+5\sqrt{13}}{9}\end{array}\right)
\left(\begin{array}{c}s\\t\end{array}\right)\\
&=&
\frac{26-5\sqrt{13}}{9}s^2+\frac{26+5\sqrt{13}}{9}t^2
\end{eqnarray*}
Since both $\frac{26-5\sqrt{13}}{9},\frac{26+5\sqrt{13}}{9}$ are positive, then
$F(x,y)>0$ as well as $x,y$ aren't zero simultaneously.
We have 
$\left(\begin{array}{c}s\\t\end{array}\right)=B^{-1}p$ 
to compute the new components of any vector $p=\left(\begin{array}{c}x\\y\end{array}\right)$.
