## Calculus (3rd Edition)

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.