Critical points and extremum of $f(x,y)=y^2-x^2+x^3+x^2y+\frac{y^3}{3}\;\;\forall\;(x,y)\in\Bbb{R}^2$ Let $f:\Bbb{R}^2\to \Bbb{R}$ be a function defined by
\begin{align}f(x,y)=y^2-x^2+x^3+x^2y+\frac{y^3}{3}\;\;\forall\;(x,y)\in\Bbb{R}^2\end{align}
$i.$ Compute the critical points of $f$
$ii.$ Does $f$ have an extremum?
My work:
\begin{align}\frac{\partial f}{\partial x}=-2x+3x^2+2xy\qquad (1)\end{align}
\begin{align}\frac{\partial f}{\partial y}=2y+x^2+y^2\qquad (2)\end{align}
At \begin{align}\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0\end{align}
we have 
\begin{align}x=\frac{1}{2}(3x^2+2xy)\end{align}
Substituting into $2$, we get 
\begin{align}0=y(8+12x^3)+y^2(4x^2+4)+9x^4\end{align}
\begin{align}y=\frac{-2-3 x^3\pm\sqrt{4+12 x^3-9 x^4}}{2 \left(1+x^2\right)}\end{align}
I don't know where to go from here. Can someone please, help?

\begin{align}\frac{\partial f}{\partial x}=-2x+3x^2+2xy\qquad (1)\end{align}
\begin{align}\frac{\partial f}{\partial y}=2y+x^2+y^2\qquad (2)\end{align}
At \begin{align}\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0\end{align}
we have (1) implies
\begin{align}x(3x+2y-2)=0\end{align}
So there are two cases $x=0$ or ... continue from there.
