\(\log_x\dfrac{1}{3}\geq2\\ x^2\leq\dfrac{1}{3}\\ x\leq \dfrac{\sqrt3}{3} \;or\;x\leq-\dfrac{\sqrt3}{3}\\ \therefore x\leq \dfrac{\sqrt3}{3}\\(\text{because the range }x\leq-\dfrac{\sqrt3}{3}\text{ coincides with the range }x\leq \dfrac{\sqrt3}{3})\)
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