# Solving $4y^4 - 4x^4 + x + y = 0$ (equation system of partial derivates)

I need help solving the following equation system: $$\frac{\partial}{\partial x} = 8xy + 4y^2 + \frac{y}{x^2 + y^2} = 0$$

$$\frac{\partial}{\partial y} = 8xy + 4x^2 - \frac{x}{x^2 + y^2} = 0$$

I've managed to simplify this down to the following: $$4x^4 - 4y^4 + x + y \Leftrightarrow 4x^4 + x = 4y^4 - y$$

But I can't find a way to separate the $x$ or $y$. Wolfram-Alpha says the solution is $y = -x$, and I'm assuming there must be some simple trick (creating a square etc) to find that solution since I very much doubt solving a general 4th degree equation is part of this particular problem, but I just can't figure out how.

The full problem is from an exercise in my course book in multivariable calculus, I'll post the whole problem below if it turns out I missed something essential in the problem statement.

Please help, it's extremely frustrating to get stuck on this subproblem which really has nothing to do with calculus at all. If there is some trick that can be applied to solve this particular equation I very much would like to know and never get stuck like this again.

Whole problem:

"Find and classify the stationary points of the following function" $$f(x,y) = 4xy(x+y) + \arctan \frac{x}{y}$$

Multiply each side of the first equation by $y$, each side of the second equation by $x$, and add. We get $8x^2y+4xy^2+8xy^2+4x^2y=0$, that is, $12xy(x+y)=0$.
From this we get that $x=-y$ or $x=0$ or $y=0$. Substitute in the first equation to finish.
Write your system of equations as $$- 8 x y = 4 y^2 + \dfrac{y}{x^2 + y^2} = 4 x^2 - \dfrac{x}{x^2 + y^2}$$ Then $$4 x^2 - 4 y^2 = 4 (x+y)(x-y) = \dfrac{x+y}{x^2 + y^2}$$ Since $x+y$ is a factor of both sides, one possibility is $x+y=0$. The other is $4 (x-y) = 1/(x^2 + y^2)$, in which case your first equation becomes $-8xy = 4 y^2 + 4 (x-y) y = 4 x y$. But then $x = 0$ or $y = 0$...