1. The problem is to verify the domain of the function $y = \frac{\sqrt{3x-4}}{x}$ given as $\left(\frac{4}{3}, \infty\right)$.\n\n2. The domain of a function is the set of all $x$ values for which the function is defined. For this function, two conditions must be met:\n - The expression inside the square root must be non-negative: $3x - 4 \geq 0$.\n - The denominator $x$ cannot be zero.\n\n3. Solve the inequality for the radicand:\n$$3x - 4 \geq 0 \implies 3x \geq 4 \implies x \geq \frac{4}{3}.$$\n\n4. The denominator $x$ cannot be zero, so $x \neq 0$. Since $\frac{4}{3} > 0$, this condition is automatically satisfied for $x \geq \frac{4}{3}$.\n\n5. Therefore, the domain is all $x$ such that $x \geq \frac{4}{3}$, or in interval notation, $\left[\frac{4}{3}, \infty\right)$.\n\n6. The given domain is $\left(\frac{4}{3}, \infty\right)$, which excludes $\frac{4}{3}$. However, at $x=\frac{4}{3}$, the radicand is zero and the denominator is nonzero, so the function is defined and equals zero.\n\n7. Hence, the correct domain should include $\frac{4}{3}$: $\left[\frac{4}{3}, \infty\right)$.\n\nFinal answer: The domain is $\boxed{\left[\frac{4}{3}, \infty\right)}$, so the given domain is almost correct but should include $\frac{4}{3}$.