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.

It is easy to estimate for instance:
$$
\begin{aligned}
|x^5-4x^3y^2-xy^4|
&\le |x^5|+|-4x^3y^2|+|-xy^4| \\
&= |x|\cdot( x^4+4x^2y^2+y^4) \\
&\le |x|\cdot( \ (x^2+y^2)^2 + 2(x^2+y^2)^2 + (x^2+y^2)^2\ )\\
&= 4|x|\; (x^2+y^2)^2\ .
\end{aligned}
$$
So is $f$ denotes the given function we have $|f(x,y)|\le 4|x|$, which holds also in zero (which is $(0,0)$). This estimation leads to continuity in zero.
