How can I show that $f(x,y)=(x^2+y^2,x^2-y^2)$ is continous at $(x_0,y_0)$ by the precise definition of the limit ($\epsilon$-$\delta$ definition)?
Fix $\epsilon>0$. We need to find $\delta>0$ (in terms of $\epsilon$) such that for all $(x,y)\in\mathbb{R}^2$ $$ 0<|(x,y)-(x_0,y_0)|<\delta\Longrightarrow|(x^2+y^2,x^2-y^2)-(x_0^2+y_0^2,x_0^2-y_0^2)|<\epsilon $$ I can see that $$ |(x,y)-(x_0,y_0)|=|(x-x_0,y-y_0)|=\sqrt{(x-x_0)^2+(y-y_0)^2} $$ and that $$ \begin{align} |(x^2+y^2,x^2-y^2)-(x_0^2+y_0^2,x_0^2-y_0^2)|&=|(x^2+y^2-x_0^2-y_0^2,x^2-y^2-x_0^2+y_0^2)|\\ &=\sqrt{(x^2+y^2-x_0^2-y_0^2)^2+(x^2-y^2-x_0^2+y_0^2)^2}\\ &=\sqrt{2}\sqrt{(x^2-x_0^2)^2+(y^2-y_0^2)^2}\\ &=\sqrt{2}\sqrt{(x-x_0)^2(x+x_0)^2+(y-y_0)^2(y+y_0)^2} \end{align} $$ and then I am not sure what to do next as the expression somehow looks messy.
For $|(x,y)-(x_{0},y_{0})|<\delta=\min\{1,\epsilon\}$, then $|x-x_{0}|<1$ and $|y-y_{0}|<1$, so $|x+x_{0}|\leq|x-x_{0}|+2|x_{0}|<1+2|x_{0}|$, similar to $|y+y_{0}|<1+2|y_{0}|$, then \begin{align*} &\sqrt{|x-x_{0}|^{2}|x+x_{0}|^{2}+|y-y_{0}|^{2}|y+y_{0}|^{2}}\\ &\leq\sqrt{1+2|x_{0}|+1+2|y_{0}|}\sqrt{|x-x_{0}|^{2}+|y-y_{0}|^{2}}\\ &=\sqrt{2}\sqrt{1+2|x_{0}|+2|y_{0}|}\sqrt{|x-x_{0}|^{2}+|y-y_{0}|^{2}}\\ &<\sqrt{2}\sqrt{1+2|x_{0}|+2|y_{0}|}\delta, \end{align*} the rest should be easy.