<s>[INST] Show that $-\frac{2yx^3}{(x^2+y^2)^2}$ is bounded. 
Show that $-\frac{2yx^3}{(x^2+y^2)^2}$ is bounded.

I'm approaching this starting with
$$ \left| \frac{2yx^3}{(x^2+y^2)^2} \right| = \left| \frac{2yx^3}{x^4+2x^2y^2+y^4} \right| \leq \left| \frac{2yx^3}{2x^2y^2} \right| = \left| \frac{x}{y}\right|.$$
However, this doesn't get me anywhere. What am I missing?
 [/INST] By AM-GM,
$$ |yx^3| = 27\left|y\cdot\frac{x}{3}\cdot\frac{x}{3}\cdot\frac{x}{3}\right| \leq 27\left(\frac{|x|+|y|}{4}\right)^4 = \frac{27}{64}(|x|+|y|)^4$$
while by AM-QM:
$$ \frac{x^2+y^2}{2} \geq \left(\frac{|x|+|y|}{2}\right)^2 $$
hence:
$$\left|\frac{2yx^3}{(x^2+y^2)^2}\right|\leq \frac{\frac{27}{32}}{\frac{1}{4}}=\frac{27}{8}.$$
An even sharper bound (thanks to Git Gud) is:
$$\left|\frac{2yx^3}{(x^2+y^2)^2}\right|\leq \left|\frac{2xy}{x^2+y^2}\right|\cdot \left|\frac{x^2}{x^2+y^2}\right|\leq 1,$$
but the sharpest possible bound is given by $\frac{3\sqrt{3}}{8}$, as shown in the comments.
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