#### Answer

See the proof below.

#### Work Step by Step

Along the lines $ y=mx $, we have
$$ f(x,y)=\frac{x^3+y^3}{x y^2}=\frac{x^3+m^3x^3}{m^2 x^3}=\frac{1+m^3}{m^2}.$$
Hence, along the paths $ y=mx $, we have
$$ \lim\limits_{(x,y) \to (0,0)}\frac{x^3+y^3}{x y^2}= \lim\limits_{(x,y) \to (0,0)}\frac{xy}{x^2+y^2}=\frac{1+m^3}{m^2}$$
hence the limit depends on the slope $ m $ and so it does not exist.