To find the length of the curve given by the equation:

$x = \frac{y^3}{4} + \frac{1}{3y}$

with the range $1 \leq y \leq 3$, we use the formula for the length of a curve $L$ given by:

$L = \int_{a}^{b} \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy$

Step 1: Find $\frac{dx}{dy}$

Given the equation:

$x = \frac{y^3}{4} + \frac{1}{3y}$

Let's differentiate $x$ with respect to $y$:

$\frac{dx}{dy} = \frac{d}{dy} \left( \frac{y^3}{4} + \frac{1}{3y} \right)$

Differentiating term by term:

$\frac{d}{dy} \left( \frac{y^3}{4} \right) = \frac{3y^2}{4}$

$\frac{d}{dy} \left( \frac{1}{3y} \right) = -\frac{1}{3y^2}$

Thus:

$\frac{dx}{dy} = \frac{3y^2}{4} - \frac{1}{3y^2}$

Step 2: Set up the integral

Now, the curve length $L$ is given by:

$L = \int_{1}^{3} \sqrt{1 + \left( \frac{3y^2}{4} - \frac{1}{3y^2} \right)^2} \, dy$

Step 3: Simplify the integrand

Simplifying the expression inside the square root:

$\left( \frac{3y^2}{4} - \frac{1}{3y^2} \right)^2 = \left( \frac{9y^4}{16} - \frac{2}{12} + \frac{1}{9y^4} \right)$

$= \frac{9y^4}{16} - \frac{1}{6} + \frac{1}{9y^4}$

Thus, the integral becomes:

$L = \int_{1}^{3} \sqrt{1 + \frac{9y^4}{16} - \frac{1}{6} + \frac{1}{9y^4}} \, dy$

This integral might be complex to solve by hand, so numerical methods or a calculator would typically be used to compute it. Would you like me to estimate the result for you using a numerical approach?