Find stationary points of the function $f(x,y) = (y^2-x^4)(x^2+y^2-20)$ I have problem in finding some of the stationary points of the function above. I proceeded in this way: the gradient of the function is:
$$ \nabla f = \left( xy^2-3x^5-2x^3y^2+40x^3 ; x^2y+2y^3-x^4y-20y \right) $$
So in order to find the stationary points, I must resolve the system:
$$ \begin{cases} 
xy^2-3x^5-2x^3y^2+40x^3 = 0 \\
x^2y+2y^3-x^4y-20y = 0
\end{cases} $$
So far I've found the points:
$$ (0,0) \qquad \left(\pm 2 \sqrt{10 \over 3} , 0 \right) \qquad (0, \pm \sqrt{10}) $$
But, I'm still blocked  when I've to found the points deriving by the system:
$$
\begin{cases} 
2x^6 + 3x^4 +x^2 -20 = 0 \\
y^2 = \frac{1}{2} \left( x^4 - x^2 + 20 \right) \end{cases}
$$
Which I don't know how to solve. Can someone help me ? Thanks.

As you implied, you can factor the original system as $x(3x^4-40x^2+2x^2y^2-y^2)$ and $y(x^4-x^2-2y^2+20)=0$.
Taking $y=0$ gives you $x=0,x=\pm2\sqrt{\frac{10}{3}}$. Taking $x=0$ gives you $y=\pm\sqrt{10}$. 
So you are left to solve $3x^4-40x^2+2x^2y^2-y^2=0,x^4-x^2-2y^2+20=0$ and $x,y\ne0$. Substituting from the second into the first gives $2x^6+3x^4-39x^2-20=0$ (you dropped the $-40x^2$ here), which factorises as: $$(x-2)(x+2)(x^2+5)(2x^2+1)$$
Can I leave you to finish?
