1. **State the problem:** Given that $xy < 0$, $$\frac{1}{x^2} + \frac{1}{y^2} = 40,$$ and $$x + y = \frac{1}{3},$$ find the value of $$\frac{1}{x^4} + \frac{1}{y^4}.$$\n\n2. **Express given terms with symmetric sums:** We want to find $$\frac{1}{x^4} + \frac{1}{y^4}.$$ Note that $$\frac{1}{x^2} + \frac{1}{y^2} = 40.$$\n\n3. **Relate $$\frac{1}{x^2} + \frac{1}{y^2}$$ to $$\frac{1}{x} + \frac{1}{y}$$ and $$\frac{1}{xy}$$:** Use the identity $$\left(\frac{1}{x} + \frac{1}{y}\right)^2 = \frac{1}{x^2} + 2\frac{1}{xy} + \frac{1}{y^2}.$$\n\n4. **Find $$\frac{1}{x} + \frac{1}{y}$$:** Since $$x + y = \frac{1}{3},$$ and $$xy = p$$ (unknown for now), then $$\frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy} = \frac{\frac{1}{3}}{p} = \frac{1}{3p}.$$\n\n5. **Express $$\frac{1}{x^2} + \frac{1}{y^2}$$ in terms of $$p$$:** Using step 3,\n$$\frac{1}{x^2} + \frac{1}{y^2} = \left(\frac{1}{x} + \frac{1}{y}\right)^2 - 2\frac{1}{xy} = \left(\frac{1}{3p}\right)^2 - 2 \cdot \frac{1}{p} = \frac{1}{9p^2} - \frac{2}{p}.$$\n\n6. **Set equal to 40 and solve for $$p$$:**\n$$\frac{1}{9p^2} - \frac{2}{p} = 40.$$ Multiply both sides by $$9p^2$$:\n$$1 - 18p = 360p^2.$$\nRearranged:\n$$360p^2 + 18p - 1 = 0.$$\n\n7. **Solve the quadratic equation:**\nUse formula $$p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=360, b=18, c=-1.$$\nCalculate discriminant:\n$$b^2 - 4ac = 18^2 - 4\times 360 \times (-1) = 324 + 1440 = 1764.$$\nSquare root:\n$$\sqrt{1764} = 42.$$\n\nTherefore,\n$$p = \frac{-18 \pm 42}{720}.$$\nTwo possible values:\n$$p_1 = \frac{24}{720} = \frac{1}{30}, \quad p_2 = \frac{-60}{720} = -\frac{1}{12}.$$\n\nSince $$xy = p$$ and $$xy < 0$$ given, choose $$p = -\frac{1}{12}.$$\n\n8. **Calculate $$\frac{1}{x} + \frac{1}{y}$$:**\n$$\frac{1}{x} + \frac{1}{y} = \frac{1}{3p} = \frac{1}{3 \times -\frac{1}{12}} = \frac{1}{-\frac{1}{4}} = -4.$$\n\n9. **Find the desired value $$\frac{1}{x^4} + \frac{1}{y^4}$$:**\nNote the identity $$\left(\frac{1}{x^2} + \frac{1}{y^2}\right)^2 = \frac{1}{x^4} + 2 \frac{1}{x^2 y^2} + \frac{1}{y^4}.$$\nRearranged,\n$$\frac{1}{x^4} + \frac{1}{y^4} = \left(\frac{1}{x^2} + \frac{1}{y^2}\right)^2 - 2 \frac{1}{x^2 y^2}.$$\n\n10. **Calculate $$\frac{1}{x^2 y^2}$$:**\n$$xy = -\frac{1}{12} \implies (xy)^2 = \frac{1}{144}.$$ So, $$\frac{1}{x^2 y^2} = \frac{1}{(xy)^2} = 144.$$\n\n11. **Calculate $$\left(\frac{1}{x^2} + \frac{1}{y^2}\right)^2$$:**\nGiven $$\frac{1}{x^2} + \frac{1}{y^2} = 40,$$ so\n$$40^2 = 1600.$$\n\n12. **Final answer:**\n$$\frac{1}{x^4} + \frac{1}{y^4} = 1600 - 2 \times 144 = 1600 - 288 = 1312.$$\n\nHence, the value is **1312**.