Continuity of $\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ at (0, 0) I am having trouble proving that $\dfrac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ is continuous at $(0, 0)$ if we set the value at $(0, 0)$ to be $0$.
I don't see a way to prove this as I cannot factor this into partial fractions.

As an another approach:
$$\bigg|\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}\bigg|=|x|\bigg|\frac{x^4-4x^2y^2-y^4}{(x^2+y^2)^2}\bigg|=|x|\bigg|\frac{(x^2+y^2)^2-6x^2y^2-2y^4}{(x^2+y^2)^2}\bigg| \\ \le |x|\bigg(1+\bigg|\frac{6x^2y^2+2y^4}{(x^2+y^2)^2}\bigg|\bigg)\le |x|\bigg(1+\bigg|\frac{6x^2y^2+6y^4}{(x^2+y^2)^2}\bigg|\bigg) \\ =|x|\bigg(1+\bigg|\frac{6y^2}{x^2+y^2}\bigg|\bigg)\le |x|(1+6)=7|x|.$$
Therefore, the limit is zero as $x$ approaches zero.
