Domain of Function: Finding Valid Inputs for y=-(x-1/3)²

The given function is $y = -\left(x - \frac{1}{3}\right)^2$. This function is in the vertex form of a quadratic equation, where the vertex is at $x = \frac{1}{3}$. The presence of the negative sign in front of the squared term indicates that the parabola opens downward.

Let's analyze the function domain in terms of where it is positive or negative:

• Since the parabola opens downward, the maximum value at the vertex $x = \frac{1}{3}$ is zero.
• The function value is negative for every $x \neq \frac{1}{3}$.
• For $x < 0$, the function will be negative because the shape of the parabola ensures negativity on either side of the vertex.
• Similarly, for $x > 0$, the function will also be negative.

Thus, for the domain where $f(x) < 0$, we have $x < 0 : x \neq \frac{1}{3}$ (though technically beyond zero these negatives are in the left half, described succinctly under real numbers as $-\infty < x < \frac{1}{3}$), where no exceptions apply for $x > 0$ in theoretical range as the parabola negatively surpasses all specified x-real, non-zero magnitude cuts off considered here by environment instruction formats.

Considering these factors, the function is never positive for any $x$, conforming exactly to the negative streak overlook starting from any negative below boundary tunneled vertex center exclusion applicability.

The correct choice based on the given options is:

• Positive Domain: None
• Negative Domain: $x < 0 : x \neq \frac{1}{3}$

Therefore, the solution is best depicted by choice 3:

$x < 0 : x\ne\frac{1}{3}$

$x > 0 :$ none

Hence, the correct answer matching the function characteristics is:

$x < 0 : x\ne\frac{1}{3}$

$x > 0 :$ none