As suggested, let's use polar coordinates:
[tex]\begin{cases}x=\rho\cos(\theta)\\y=\rho\sin(\theta)\end{cases}[/tex]
To get
[tex]f(x,y)=\dfrac{x^3-xy^2}{x^2+y^2}\mapsto \dfrac{\rho^3\cos^3(\theta)-\rho^3\cos(\theta)\sin^2(\theta)}{\rho^2}[/tex]
We can simplify the expression to
[tex]\dfrac{\rho^3\cos^3(\theta)-\rho^3\cos(\theta)\sin^2(\theta)}{\rho^2}=\dfrac{\rho^3\cos(\theta)(\cos^2(\theta)-\sin^2(\theta))}{\rho^2}[/tex]
And simplify [tex]\rho^2[/tex] to get
[tex]\rho\cos(\theta)(\cos^2(\theta)-\sin^2(\theta))[/tex]
So, as [tex]\rho\to 0[/tex], [tex]f(x,y)\to 0[/tex] as well.
