1. **Problem:** Simplify $\sqrt[3]{\frac{49}{27}}\,j^{1/3}$.\n2. Use the rule $a^{1/3}=\sqrt[3]{a}$. So $j^{1/3}=\sqrt[3]{j}$.\n3. Rewrite the expression as $\sqrt[3]{\frac{49}{27}}\,\sqrt[3]{j}=\sqrt[3]{\frac{49j}{27}}$.\n4. Now simplify the denominator inside the cube root by separating perfect cubes: $27=3^3$.\n5. So we can write $$\sqrt[3]{\frac{49j}{27}}=\frac{\sqrt[3]{49j}}{\sqrt[3]{27}}.$$\n6. Evaluate the cube root of $27$: $$\sqrt[3]{27}=3.$$\n7. Therefore the simplified form is $$\frac{\sqrt[3]{49j}}{3}.$$\n8. **Final answer:** $\frac{\sqrt[3]{49j}}{3}$