How can I solve the system of equations? How can I solve the system of equations? 
$$\begin{cases}
x^2 y^2+12 x y^3-18 x y-18y^4-4 y^2+27=0,&\\
x^2 y^2-3 x y^3-3 x y+5 y^2=0.
\end{cases}$$
I have not any idea to solve.

$$\begin{cases}
x^2 y^2+12 x y^3-18 x y-18y^4-4 y^2+27=0,&\\
x^2 y^2-3 x y^3-3 x y+5 y^2=0.
\end{cases}$$
We subtract the second equation from the first to get 
$15xy^3-15xy-18y^4-9y^2+27=0$. 
But $15xy^3-15xy=15xy(y^2-1)$, and $-18y^4-9y^2+27=-9(2y^4+y^2-3)=9(-2y^2-3)(y^2-1)$. 
Hence $15xy^3-15xy-18y^4-9y^2+27$$=(y^2-1)(15xy-18y^2-27)=3(y^2-1)(5xy-6y^2-9)$. Thus either $y^2=1$ (hence $y=\pm 1$, or $5xy-6y^2-9=0$.
Case 1: $y=\pm 1$. 
If $y=1$, Substitute to the second equation to get $x^2-6x+5=0$, so $x=1$ or $x=5$. 
If $y=-1$, Substitute to the second equation to get $x^2+6x+5=0$, so $x=-1$ or $x=-5$. 
We can get these solutions by factoring. 
Case 2: $5xy-6y^2-9=0$
So now we proceed to write $x^2 y^2-3 x y^3-3 x y+5 y^2=0$ in terms of $y$, by changing all $xy$ terms into $\frac{6y^2-9}{5}$. 
Thus $x^2y^2-3xy^3-3xy+5y^2=(\frac{6y^2-9}{5})^2-(\frac{6y^2-9}{5})y^2-3(\frac{6y^2-9}{5})+5y^2=0$. This is just a quadratic in $y^2$. But then this quadratic is equal to $3(y^2)^2-14(y^2)+108=0$ (after expanding and regrouping), and it has no real roots as its discriminant, $14^2-4(3)(108)$, is negative. 
Thus our only solutions are $\boxed{(x,y)=(\pm 1, \pm 5), (\pm 5, \pm1)}$. 
