How this type of equation is solved?

Question

I'm solving a relative and ends when the function (x²+y²)²-4x² derive out this equation, but not that I have to do to get resolve it

$$f'x = 4x³+4y²x-8x$$

$$f'y = (4x²+4y⁴)y$$

How solve this ecuation system?

$$4x³+4y²x-8x=0$$

$$(4x²+4y⁴)y=0$$

Answer 1

You can take $(4x^2+4y^4)y=0$ and first say $y=0$ is a solution, then plug that into the first and find $0=x^3-2x=x(x^2-2)$ and the solutions are $0, \pm \sqrt 2$.
If $y \neq 0$, then $x^2=y^4, x=\pm y^2$ and the first becomes (the $\pm$ divides out) $x^3+x^2-2x=0$ or $x(x^2+x-2)=0$, the second part of which is a quadratic.