Proving $\lim_{(x, y) \rightarrow (0, 0)} \frac{x^4 - y^4}{x^3 - y^3} = 0$ [closed]

Let $f(x, y) = \dfrac{x^4 - y^4}{x^3 - y^3}$ be defined in $U = \mathbb{R}^2 \setminus \{(x, x) \colon x \in \mathbb{R}\}$. I have to determine $\lim_{(x, y) \rightarrow (0, 0)} f(x, y)$ by $\varepsilon$-$\delta$.

I know that if $(x, y) \in U$, then $\dfrac{x^4 - y^4}{x^3 - y^3} = \frac{(x - y)(x + y)(x^2 + y^2)}{(x - y)(x^2 + xy + y^2)} = \frac{(x + y)(x^2 + y^2)}{x^2 + xy + y^2}.$Now I got stuck...