Simultaneous real solution of $x^3+y^3+1+6xy=0$ & $xy^2+y+x^2=0$ I am trying to solve the following system of non-linear equations in real numbers: 
$x^3+y^3+1+6xy=0$ & $xy^2+y+x^2=0$, with $x,y$ real.
I can only see that $xy\ne 0$. 
I have no clue whether a solution exists or not and how to find any solution. I cannot seem to be able to separate $x,y$ or write it in a good parametric form.
Please help.

Not a very satisfactory answer since it involves the cubic formula...
Set $u = xy$. Multiply the first equation by $x^3$ and the second by $x$. Then we get 
$x^6+u^3+x^3+6ux^3 = 0$
$u^2+u+x^3 = 0$
Solving for $x^3$ in the second and substituting into the first, we get 
$u^4+2u^3+u^2+u^3-u^2-u+6u(-u^2-u)=0$
$u^4+2u^3+u^2+u^3-u^2-u-6u^3-6u^2=0$
$u^4-3u^3-6u^2-u=0$
$u^3-3u^2-6u-1=0$
Which can be solved by the cubic formula. Once this is done, $u$ can be substituted back in to find $x$. Then we can find $y$.
