Solution:

Let $x = \cos(\theta)$ and $y = \sin(\theta)$. Then, by the triple angle formulae, we have that $3x - 4x^{3} = -\cos(3\theta)$ and $3y - 4y^{3} = \sin(3\theta)$, so $-\sin(3\theta) \cos(3\theta) = -\frac{1}{2}$. We can write this as $2 \sin(3\theta) \cos(3\theta) = \sin(6\theta) = 1$, so $\theta = \frac{1}{6} \sin^{-1}(1) = \frac{\pi}{12}$. Thus,

$x + y = \cos\left(\frac{\pi}{12}\right) + \sin\left(\frac{\pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4} + \frac{\sqrt{6} - \sqrt{2}}{4} = \frac{\sqrt{6}}{2}$.

Solution 2:

Expanding gives $9xy + 16x^{3}y^{3} - 12x y^{3} - 12x^{3}y = 9(xy) + 16(xy)^{3} - 12(xy)(x^{2} + y^{2}) = -\frac{1}{2}$, and since $x^{2} + y^{2} = 1$, this is $-3(xy) + 16(xy)^{3} = -\frac{1}{2}$, giving $xy = -\frac{1}{2}, \frac{1}{4}$. However, since $x$ and $y$ are positive reals, we must have $xy = \frac{1}{4}$. Then,

$x + y = \sqrt{x^{2} + y^{2} + 2xy} = \sqrt{1 + 2 \cdot \frac{1}{4}} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$.