1. **State the problem:** Differentiate the function $$f(x) = \frac{\sqrt{3+x}}{\sqrt[3]{x^2-2}}$$ with respect to $$x$$. 2. **Rewrite the function using exponents:** $$f(x) = (3+x)^{\frac{1}{2}} \cdot (x^2 - 2)^{-\frac{1}{3}}$$ 3. **Apply the product rule:** If $$f(x) = u(x) \cdot v(x)$$, then $$f'(x) = u'(x)v(x) + u(x)v'(x)$$. Here, $$u(x) = (3+x)^{\frac{1}{2}}$$ and $$v(x) = (x^2 - 2)^{-\frac{1}{3}}$$. 4. **Find $$u'(x)$$:** Using the chain rule, $$u'(x) = \frac{1}{2}(3+x)^{-\frac{1}{2}} \cdot 1 = \frac{1}{2}(3+x)^{-\frac{1}{2}}$$ 5. **Find $$v'(x)$$:** Using the chain rule, $$v'(x) = -\frac{1}{3}(x^2 - 2)^{-\frac{4}{3}} \cdot 2x = -\frac{2x}{3}(x^2 - 2)^{-\frac{4}{3}}$$ 6. **Combine using the product rule:** $$f'(x) = \frac{1}{2}(3+x)^{-\frac{1}{2}} (x^2 - 2)^{-\frac{1}{3}} + (3+x)^{\frac{1}{2}} \left(-\frac{2x}{3}(x^2 - 2)^{-\frac{4}{3}}\right)$$ 7. **Simplify the expression:** $$f'(x) = \frac{1}{2}(3+x)^{-\frac{1}{2}} (x^2 - 2)^{-\frac{1}{3}} - \frac{2x}{3}(3+x)^{\frac{1}{2}} (x^2 - 2)^{-\frac{4}{3}}$$ This is the derivative of the given function. **Final answer:** $$\boxed{f'(x) = \frac{1}{2}(3+x)^{-\frac{1}{2}} (x^2 - 2)^{-\frac{1}{3}} - \frac{2x}{3}(3+x)^{\frac{1}{2}} (x^2 - 2)^{-\frac{4}{3}}}$$