Show $f(x,y)=\frac{x^3\sin(x-y)}{x^3-y^3}$ for $x\neq y$, $f(x,y)=0$ for $x=y$ is continuous at $(0,0)$. I know that once we bound $f$ for $x\neq y$ we are done because $f=0$ for $x=y$. I got that for $x\neq y$, $|f(x,y)|\le \left|\frac{x^3}{x^2+xy+y^2}\right|$ because $\left|\frac{\sin(a)}{a}\right|\leq 1$ for $a\neq 0$. But I don't know how to continue.

so following @Kelenner's hint, $x^2+xy+y^2 = \frac{x^2+y^2}{2} +\frac{(x+y)^2}{2} \geq \frac{x^2+y^2}{2}$, in particular it must be positive, and $|f(x,y)|\leq |\frac{x^3}{x^2+xy+y^2}| \leq |\frac{2x^3}{x^2+y^2}| \leq |\frac{2x^3}{x^2}| =2|x|$, which tends to $0$ as $(x,y)\rightarrow(0,0)$.
We can also say that $|f(x,y)|\leq |\frac{x^3}{x^2+xy+y^2}| \leq |\frac{2x^3}{x^2+y^2}| \leq \frac{2x^2(x^2+y^2)^{\frac{1}{2}}}{x^2+y^2} \leq 2x^2(x^2+y^2)^{\frac{1}{2}}$ which also tends to $0$.
