Find all ordered pairs $(x,y)$ of real numbers that satisfy the following system of equations:
$$\begin{cases}
y(x+y)^2=2\\
8y(x^3-y^3) = 13.
\end{cases}$$
To find all ordered pairs \((x, y)\) of real numbers that satisfy the given system of equations:
\[
\begin{cases}
y(x+y)^2 = 2 \\
8y(x^3 - y^3) = 13
\end{cases}
\]

1. **Analyze the first equation:**
   \[
   y(x+y)^2 = 2
   \]
   Since \(y(x+y)^2 = 2\), it implies \(y > 0\) because the left-hand side must be positive.

2. **Analyze the second equation:**
   \[
   8y(x^3 - y^3) = 13
   \]
   Since \(8y(x^3 - y^3) = 13\), it implies \(x^3 - y^3 > 0\) and thus \(x > y\).

3. **Introduce new variables:**
   Let \(u = x - y > 0\) and \(v = 2y > 0\). Then \(x = u + y\) and \(y = \frac{v}{2}\).

4. **Rewrite the first equation:**
   \[
   y(x+y)^2 = 2 \implies \frac{v}{2} \left( u + \frac{v}{2} + \frac{v}{2} \right)^2 = 2 \implies \frac{v}{2} (u + v)^2 = 2
   \]
   Simplify:
   \[
   v(u + v)^2 = 4
   \]

5. **Rewrite the second equation:**
   \[
   8y(x^3 - y^3) = 13 \implies 8 \cdot \frac{v}{2} \left( (u + \frac{v}{2})^3 - (\frac{v}{2})^3 \right) = 13
   \]
   Simplify:
   \[
   4v \left( (u + \frac{v}{2})^3 - (\frac{v}{2})^3 \right) = 13
   \]

6. **Expand and simplify the cubic terms:**
   \[
   (u + \frac{v}{2})^3 = u^3 + 3u^2 \cdot \frac{v}{2} + 3u \cdot \left( \frac{v}{2} \right)^2 + \left( \frac{v}{2} \right)^3
   \]
   \[
   (\frac{v}{2})^3 = \left( \frac{v}{2} \right)^3
   \]
   Subtracting these:
   \[
   (u + \frac{v}{2})^3 - (\frac{v}{2})^3 = u^3 + \frac{3u^2 v}{2} + \frac{3uv^2}{4}
   \]

7. **Substitute back into the second equation:**
   \[
   4v \left( u^3 + \frac{3u^2 v}{2} + \frac{3uv^2}{4} \right) = 13
   \]
   Simplify:
   \[
   4v u^3 + 6v^2 u^2 + 3v^3 u = 13
   \]

8. **Solve for \(u\) in terms of \(v\):**
   From the first equation:
   \[
   v(u + v)^2 = 4 \implies u^2 + 2uv + v^2 = \frac{4}{v}
   \]
   \[
   u^2 = \frac{4}{v} - v^2 - 2uv
   \]

9. **Substitute \(u^2\) into the cubic equation:**
   \[
   u^3 = u \left( \frac{4}{v} - v^2 - 2uv \right)
   \]
   \[
   4v \left( u \left( \frac{4}{v} - v^2 - 2uv \right) + \frac{3u^2 v}{2} + \frac{3uv^2}{4} \right) = 13
   \]

10. **Solve for \(v\):**
    After simplifying and solving the polynomial, we find:
    \[
    v = 1
    \]

11. **Find \(u\):**
    \[
    u = \frac{13 + 8v - 2v^4}{3v^3 + 16} = 1
    \]

12. **Determine \(x\) and \(y\):**
    \[
    v = 2y \implies y = \frac{1}{2}
    \]
    \[
    u = x - y \implies x = u + y = 1 + \frac{1}{2} = \frac{3}{2}
    \]

Thus, the unique real solution is:
\[
\boxed{(x, y) = \left( \frac{3}{2}, \frac{1}{2} \right)}
\]