Uniform continuity of $f(x)=x^{2/3}\log x$ on $[0, \infty)$
Solution 1:
You proved that the function $f$ is continuous on $[0,1]$, hence uniformly continuous on $[0,1]$. The expression of $f'$ you provided, plus the inequality $\log u\leqslant u$ applied to $u=x^{1/3}$, yield $0\leqslant f'(x)\leqslant3$ for every $x\geqslant1$, hence $f$ is uniformly continuous on $[1,+\infty)$. Ergo.