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