1. **State the problem:** Find the first derivative $f'(x)$ and the second derivative $f''(x)$ of the function $$f(x) = \frac{x}{x^2 - 2}.$$\n\n2. **Recall the formula:** For a function given as a quotient $f(x) = \frac{u(x)}{v(x)}$, the derivative is found using the quotient rule: $$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}.$$\n\n3. **Identify $u(x)$ and $v(x)$:** Here, $u(x) = x$ and $v(x) = x^2 - 2$.\n\n4. **Compute derivatives of numerator and denominator:** $$u'(x) = 1,$$ $$v'(x) = 2x.$$\n\n5. **Apply the quotient rule:** $$f'(x) = \frac{1 \cdot (x^2 - 2) - x \cdot 2x}{(x^2 - 2)^2} = \frac{x^2 - 2 - 2x^2}{(x^2 - 2)^2}.$$\n\n6. **Simplify the numerator:** $$x^2 - 2 - 2x^2 = -x^2 - 2.$$\n\n7. **Write the simplified first derivative:** $$f'(x) = \frac{-x^2 - 2}{(x^2 - 2)^2} = \frac{-(x^2 + 2)}{(x^2 - 2)^2}.$$\n\n8. **Find the second derivative $f''(x)$:** Differentiate $f'(x)$ using the quotient rule again, where numerator $N(x) = -(x^2 + 2)$ and denominator $D(x) = (x^2 - 2)^2$.\n\n9. **Compute derivatives:** $$N'(x) = -2x,$$ $$D'(x) = 2(x^2 - 2) \cdot 2x = 4x(x^2 - 2).$$\n\n10. **Apply quotient rule for $f''(x)$:** $$f''(x) = \frac{N'(x)D(x) - N(x)D'(x)}{[D(x)]^2} = \frac{-2x (x^2 - 2)^2 - (-(x^2 + 2)) 4x (x^2 - 2)}{(x^2 - 2)^4}.$$\n\n11. **Simplify numerator:** $$-2x (x^2 - 2)^2 + 4x (x^2 + 2)(x^2 - 2) = x \left[-2 (x^2 - 2)^2 + 4 (x^2 + 2)(x^2 - 2)\right].$$\n\n12. **Expand terms inside brackets:**\n$$-2 (x^2 - 2)^2 = -2 (x^4 - 4x^2 + 4) = -2x^4 + 8x^2 - 8,$$\n$$4 (x^2 + 2)(x^2 - 2) = 4 (x^4 - 4) = 4x^4 - 16.$$\n\n13. **Sum inside brackets:** $$-2x^4 + 8x^2 - 8 + 4x^4 - 16 = ( -2x^4 + 4x^4 ) + 8x^2 + (-8 -16) = 2x^4 + 8x^2 - 24.$$\n\n14. **Final numerator:** $$x (2x^4 + 8x^2 - 24) = 2x^5 + 8x^3 - 24x.$$\n\n15. **Write the second derivative:** $$f''(x) = \frac{2x^5 + 8x^3 - 24x}{(x^2 - 2)^4}.$$\n\n**Answer:**\n$$f'(x) = \frac{-(x^2 + 2)}{(x^2 - 2)^2}, \quad f''(x) = \frac{2x^5 + 8x^3 - 24x}{(x^2 - 2)^4}.$$