1. **Stating the problem:** We are given that $xy < 0$, $$\frac{1}{x^2} + \frac{1}{y^2} = 40,$$ and $$x + y = \frac{1}{3}.$$ We need to find the value of $$\frac{1}{x^4} + \frac{1}{y^4}.$$\n\n2. **Express knowns in terms of $x$ and $y$:** From the given, $$x + y = \frac{1}{3}$$ and the product $xy$ is unknown but negative.\n\n3. **Rewrite the expression for $\frac{1}{x^2} + \frac{1}{y^2}$:** We use the identity $$\frac{1}{x^2} + \frac{1}{y^2} = \frac{(x+y)^2 - 2xy}{(xy)^2}.$$ Substitute known values: $$40 = \frac{\left(\frac{1}{3}\right)^2 - 2xy}{(xy)^2} = \frac{\frac{1}{9} - 2xy}{(xy)^2}.$$\n\n4. **Set $t = xy$ and solve for $t$:** Multiply both sides by $t^2$: $$40 t^2 = \frac{1}{9} - 2t.$$ Rearrange: $$40 t^2 + 2t - \frac{1}{9} = 0.$$ Multiply both sides by 9 to clear denominator: $$360 t^2 + 18 t - 1 = 0.$$\n\n5. **Solve quadratic equation for $t$:** Using quadratic formula: $$t = \frac{-18 \pm \sqrt{18^2 - 4\times 360 \times (-1)}}{2 \times 360} = \frac{-18 \pm \sqrt{324 + 1440}}{720} = \frac{-18 \pm \sqrt{1764}}{720} = \frac{-18 \pm 42}{720}.$$ So, $$t_1 = \frac{24}{720} = \frac{1}{30}, \quad t_2 = \frac{-60}{720} = -\frac{1}{12}.$$ Given $xy < 0$, we take $t = xy = -\frac{1}{12}.$\n\n6. **Find $\frac{1}{x^4} + \frac{1}{y^4}$:** Use the identity: $$\frac{1}{x^4} + \frac{1}{y^4} = \left(\frac{1}{x^2} + \frac{1}{y^2}\right)^2 - 2 \times \frac{1}{x^2 y^2}.$$ We already know $$\frac{1}{x^2} + \frac{1}{y^2} = 40.$$ Also, $$\frac{1}{x^2 y^2} = \frac{1}{(xy)^2} = \frac{1}{\left(-\frac{1}{12}\right)^2} = \frac{1}{\frac{1}{144}} = 144.$$ Therefore,\n$$\frac{1}{x^4} + \frac{1}{y^4} = 40^2 - 2 \times 144 = 1600 - 288 = 1312.$$\n\n**Final answer:** $$\frac{1}{x^4} + \frac{1}{y^4} = 1312.$$