I am supposed to decide whether or not you can solve the equation clearly $ f(x, y)=(x^2+y^2)^2+3x^2y-y^3=0 $ for $x$ or $y$, with $x, y, \in U$, $(0, 0) \in U$ and $U$ open. When looking at the set of $x$ and $y$ solving that equation it is pretty clear that such a $U$ does not exist.
It is pretty easy to show that you can not solve for $x$, as for any circle around $(0, 0)$, if $(x, y)$ is in that circle, $(-x, y)$ is in the circle as well. ($f$ is symmetric, so $0=f(x, y=f(-x, y)$)
Now I would like to show that you can also not solve for $y$. Or in other words for any circle with an arbitrarily small radius if $(x, y)$ is in that circle I need to find $c$, such that $0=f(x, y)=f(x, cy)$, but I could not manage to find such a $c$.
Can anyone provide me with that $c$ or is there a better approach?
COMMENT.-First at all, a circle in a open neighbourhood of $(0,0)$ don't need its center be $(0,0)$, even $(0,0)$ could be out of the circle.
Whatever be the fixed value of $x$ or $y$, the resulting equation in one variable being of $4$-degree has four roots (real and/or imaginary ones).
Besides the equation being $x^4+(2y^2+3y)x^2+(y^4-y^3)=0$, we easily have $$x=\pm\sqrt{\frac{-(2y^2+3y)^2\pm\sqrt{16y^3+9y^2}}{2}}=g(y)\tag 1$$ We can verify that always $-(2y^2+3y)^2+\sqrt{16y^3+9y^2}\le-3$ (this negative maximum is reached for $y=-0.5$) which implies that the four functions $x=h(y)$ we can get with the $\pm$ signes from the initial one are not real functions.
For the possibility $y=h(x)$ we cannot do easily as for $x=g(y)$. Anyway I agree with the viewpoint of the O.P. on the impossibility but this feeling is not a proof.