Prove that $f(x)=x^4$ is continuous on $[0,\infty)$ I understand that we want $\varepsilon$ so that $|x^2+y^2||x+y||x-y|<\varepsilon$, and for $|x+y||x-y|<\varepsilon$ I can choose $\delta$ so that $\delta(2|y|+\delta)<\varepsilon$, but I don't know what to do with the $|x^2+y^2|$ part.

If $y=0$, then $|x^4-y^4|=x^4< \epsilon$ whenever $|x-y|=|x|<\epsilon^{1/4}$.  So, take $\delta=\epsilon^{1/4}$.
If $y\ne 0$, then take $|x-y|<y/2$ so that $y/2<x<3y/2$.
Then, we have
$$\begin{align}
|x^4-y^4|&=|x-y|(x+y)(x^2+y^2)\\\\
&<|x-y|\left(\frac52 y\right)\left(\frac{13}{4}y^2\right)\\\\
&=|x-y|\frac{65}{8}y^3\\\\
&<\epsilon
\end{align}$$
whenever $|x-y|<\frac{8}{65y^3}\epsilon$.  
So, let $\delta =\min\left(\frac{y}{2},\frac{8}{65y^3}\epsilon\right)$.  Then, for all $\epsilon>0$, $|x^4-y^4|<\epsilon$ for any $x>0$, 
