Prove that $\left|\frac{x^3y^3}{9x^4+y^4}\right| \le \frac{x^2+y^2}{6}$ I need to prove that: $$\left|\frac{x^3y^3}{9x^4+y^4}\right| \le \frac{x^2+y^2}{6}$$
I am new to inequalities so I only tried C-S and  AM-GM but none of those work.
Any hints on how to proceed here?.

For $xy=0$ it's obvious.
But for $xy\neq0$ by AM-GM we obtain:
$$\left|\frac{x^3y^3}{9x^4+y^4}\right|\leq\frac{|x^3y^3|}{2\sqrt{9x^4y^4}}=\frac{|xy|}{6}\leq\frac{\frac{x^2+y^2}{2}}{6}\leq\frac{x^2+y^2}{6}.$$
