Find the maximum of $x+y$ given that $x$ and $y$ are positive real numbers that satisfy \[x^3+y^3+(x+y)^3+36xy=3456.\]
1. Let \( x + y = a \) and \( xy = b \). We aim to maximize \( a \).
2. Given the equation:
   \[
   x^3 + y^3 + (x+y)^3 + 36xy = 3456
   \]
   Substitute \( x + y = a \) and \( xy = b \):
   \[
   x^3 + y^3 + a^3 + 36b = 3456
   \]
3. Using the identity \( x^3 + y^3 = (x+y)(x^2 - xy + y^2) \) and substituting \( x + y = a \) and \( xy = b \):
   \[
   x^3 + y^3 = a(a^2 - 3b)
   \]
   Therefore, the equation becomes:
   \[
   a(a^2 - 3b) + a^3 + 36b = 3456
   \]
   Simplifying:
   \[
   a^3 - 3ab + a^3 + 36b = 3456
   \]
   \[
   2a^3 - 3ab + 36b = 3456
   \]
4. To find the maximum value of \( a \), we take the derivative with respect to \( a \):
   \[
   \frac{d}{da}(2a^3 - 3ab + 36b) = 0
   \]
   \[
   6a^2 - 3b = 0
   \]
   Solving for \( b \):
   \[
   6a^2 = 3b
   \]
   \[
   b = 2a^2
   \]
5. Substitute \( b = 2a^2 \) back into the original equation:
   \[
   2a^3 - 3a(2a^2) + 36(2a^2) = 3456
   \]
   \[
   2a^3 - 6a^3 + 72a^2 = 3456
   \]
   \[
   -4a^3 + 72a^2 = 3456
   \]
   \[
   a^3 - 18a^2 + 864 = 0
   \]
6. To solve the cubic equation \( a^3 - 18a^2 + 864 = 0 \), we use the Rational Root Theorem. The possible rational roots are the divisors of 864. Checking these, we find:
   \[
   a = 12
   \]
   satisfies the equation:
   \[
   12^3 - 18 \cdot 12^2 + 864 = 0
   \]
   \[
   1728 - 2592 + 864 = 0
   \]
   \[
   0 = 0
   \]

The final answer is \( \boxed{12} \).