Use $\,\varepsilon-\delta\,$ definition of limits to show that $\lim\limits_{(x,y)\to(-1,2)}\frac{x^3+y^3}{x^2+y^2}=\frac{7}{5}$

We need to show that for every $\varepsilon >0$ there exists a $\delta >0$ such that$$\left\lvert\frac{x^3+y^3}{x^2+y^2}-\frac{7}{5}\right\rvert<\varepsilon $$whenever$$0<\sqrt{(x+1)^2+(y-2)^2}<\delta.$$

Consider\begin{align*}\left\lvert\frac{x^3+y^3}{x^2+y^2}-\frac{7}{5}\right\rvert&=\left\lvert\frac{5(x^3+y^3)-7(x^2+y^2)}{5(x^2+y^2)}\right\rvert\\&=\left\lvert\frac{x^2(5x-7)+y^2(5y-7)}{5(x^2+y^2)}\right\rvert\\&\leqslant\left\lvert\frac{5x-7}{5}\right\rvert+\left\lvert\frac{5y-7}{5} \right\rvert\end{align*}

From here I don't know how to use the inequality given for $\delta.$