Let $x,y$ be real numbers with $x \geqslant \sqrt{2021}$ such that
\[ \sqrt[3]{x+\sqrt{2021}}+\sqrt[3]{x-\sqrt{2021}} = \sqrt[3]{y}\]
Determine the set of all possible values of $y/x$.
1. Given the equation:
   \[
   \sqrt[3]{x+\sqrt{2021}} + \sqrt[3]{x-\sqrt{2021}} = \sqrt[3]{y}
   \]
   we need to determine the set of all possible values of \( \frac{y}{x} \).

2. Let \( a = \sqrt[3]{x+\sqrt{2021}} \) and \( b = \sqrt[3]{x-\sqrt{2021}} \). Then the given equation becomes:
   \[
   a + b = \sqrt[3]{y}
   \]

3. Cubing both sides of the equation \( a + b = \sqrt[3]{y} \), we get:
   \[
   (a + b)^3 = y
   \]
   Expanding the left-hand side, we have:
   \[
   a^3 + b^3 + 3ab(a + b) = y
   \]

4. Since \( a = \sqrt[3]{x+\sqrt{2021}} \) and \( b = \sqrt[3]{x-\sqrt{2021}} \), we know:
   \[
   a^3 = x + \sqrt{2021} \quad \text{and} \quad b^3 = x - \sqrt{2021}
   \]
   Therefore:
   \[
   a^3 + b^3 = (x + \sqrt{2021}) + (x - \sqrt{2021}) = 2x
   \]

5. Next, we need to find \( ab \):
   \[
   ab = \sqrt[3]{(x+\sqrt{2021})(x-\sqrt{2021})} = \sqrt[3]{x^2 - 2021}
   \]

6. Substituting \( a^3 + b^3 \) and \( ab \) back into the expanded equation, we get:
   \[
   2x + 3ab(a + b) = y
   \]
   Let \( t = \sqrt[3]{x^2 - 2021} \). Then:
   \[
   ab = t
   \]
   and:
   \[
   y = 2x + 3t \sqrt[3]{y}
   \]

7. Let \( u = \sqrt[3]{\frac{y}{x}} \). Then:
   \[
   y = u^3 x
   \]
   Substituting \( y = u^3 x \) into the equation \( y = 2x + 3t \sqrt[3]{y} \), we get:
   \[
   u^3 x = 2x + 3t u x
   \]
   Dividing both sides by \( x \):
   \[
   u^3 = 2 + 3tu
   \]

8. We need to solve for \( u \). Rearranging the equation:
   \[
   u^3 - 3tu - 2 = 0
   \]

9. Let \( t = \sqrt[3]{1 - \frac{2021}{x^2}} \). Since \( x \geq \sqrt{2021} \), we have:
   \[
   1 - \frac{2021}{x^2} \in [0, 1)
   \]
   Therefore, \( t \in [0, 1) \).

10. Consider the function \( g(u) = \frac{u^3 - 2}{3u} \). We need to find the range of \( u \) such that \( t = g(u) \).

11. For \( u > 0 \), \( g(u) \) is continuous and increasing. We have:
    \[
    g(\sqrt[3]{2}) = 0 \quad \text{and} \quad g(2) = 1
    \]
    Therefore, \( t \in [0, 1) \) implies:
    \[
    u \in [\sqrt[3]{2}, 2)
    \]

12. Since \( \frac{y}{x} = u^3 \), the possible values of \( \frac{y}{x} \) are:
    \[
    \left[\left(\sqrt[3]{2}\right)^3, 2^3\right) = [2, 8)
    \]

The final answer is \( \boxed{\frac{y}{x} \in [2, 8)} \).