To solve this, we need to recall the formula for the volume of a pyramid:

$V = \frac{1}{3} \times \text{Base Area} \times \text{Height}$

Given that the pyramid has a square base with side length $y$, the area of the base is:

$\text{Base Area} = y^2$

Therefore, the volume formula becomes:

$V = \frac{1}{3} \times y^2 \times h$

Now, solve for the height $h$:

$h = \frac{3V}{y^2}$

Thus, the expression that represents the height of the pyramid is:

$\boxed{\frac{3V}{y^2}}$

This corresponds to the first option in the provided list.

Let me know if you would like further clarification, and here are five related questions to expand your understanding:

1. What is the general formula for the volume of a pyramid with any polygonal base?
2. How does the volume formula change for a cone?
3. If the length of the base edge $y$ is doubled, how does the height change?
4. How would the formula for the height change if the pyramid had a triangular base?
5. What is the relationship between the slant height and the perpendicular height of a pyramid?

Tip: When solving for one variable in terms of others, always check for correct algebraic manipulation and units consistency.