Identify the function and its components.

The given function is $y = 8^{\frac{1}{3x}}$. This is an exponential function with base 8 and exponent $\frac{1}{3x}$.

The exponent $\frac{1}{3x}$ is defined for all real numbers $x$ except where the denominator is zero.

So, the domain of the function is $x \in (-\infty, 0) \cup (0, \infty)$.

• If $x > 0$, then $3x > 0$, so $k = \frac{1}{3x} > 0$.
• If $x < 0$, then $3x < 0$, so $k = \frac{1}{3x} < 0$. The exponent $k$ can take any real value except 0. So, $k \in (-\infty, 0) \cup (0, \infty)$.

Since the base of the exponential function is 8 (which is positive), $y = 8^k$ will always be positive. So $y > 0$.

Consider two cases for $k$:
Case 1: $k > 0$.
If $k > 0$, then $8^k > 8^0$.
Since $8^0 = 1$, this means $8^k > 1$.
So, for $k \in (0, \infty)$, the range of $8^k$ is $(1, \infty)$.

Case 2: $k < 0$.
If $k < 0$, then $0 < 8^k < 8^0$.
Since $8^0 = 1$, this means $0 < 8^k < 1$.
So, for $k \in (-\infty, 0)$, the range of $8^k$ is $(0, 1)$.

The overall range of the function $y = 8^{\frac{1}{3x}}$ is the union of the ranges from Case 1 and Case 2.
Range $= (0, 1) \cup (1, \infty)$.
This means $y$ can be any positive real number except 1.

The final answer is $(0, 1) \cup (1, \infty)$.