For every positive real pair $(x,y)$ satisfying the equation $x^3+y^4 = x^2y$, if the greatest value of $x$ is $A$, and the greatest value of $y$ is $B$, then $A/B = ?$

$ \textbf{(A)}\ \frac{2}{3} \qquad \textbf{(B)}\ \frac{512}{729} \qquad \textbf{(C)}\ \frac{729}{1024} \qquad \textbf{(D)}\ \frac{3}{4} \qquad \textbf{(E)}\ \frac{243}{256}$
1. Given the equation \( x^3 + y^4 = x^2 y \), we need to find the greatest values of \( x \) and \( y \) and then determine the ratio \( \frac{A}{B} \).

2. Rewrite the equation in a more manageable form:
   \[
   x^3 + y^4 = x^2 y
   \]
   Divide both sides by \( y^4 \):
   \[
   \left(\frac{x}{y}\right)^3 + 1 = \left(\frac{x}{y}\right)^2 \cdot \frac{1}{y}
   \]
   Let \( a = \frac{x}{y} \). Then the equation becomes:
   \[
   a^3 + 1 = a^2 \cdot \frac{1}{y}
   \]
   Solving for \( y \):
   \[
   y = \frac{a^2}{a^3 + 1}
   \]

3. To find the maximum value of \( x \), express \( x \) in terms of \( a \) and \( y \):
   \[
   x = ay = a \cdot \frac{a^2}{a^3 + 1} = \frac{a^3}{a^3 + 1}
   \]
   We need to maximize \( \frac{a^3}{a^3 + 1} \). Consider the function:
   \[
   f(a) = \frac{a^3}{a^3 + 1}
   \]
   To find the maximum, take the derivative and set it to zero:
   \[
   f'(a) = \frac{(a^3 + 1) \cdot 3a^2 - a^3 \cdot 3a^2}{(a^3 + 1)^2} = \frac{3a^2}{(a^3 + 1)^2}
   \]
   Setting \( f'(a) = 0 \):
   \[
   3a^2 = 0 \implies a = 0
   \]
   Since \( a \) must be positive, we check the boundary as \( a \to \infty \):
   \[
   \lim_{a \to \infty} \frac{a^3}{a^3 + 1} = 1
   \]
   However, this does not provide a finite maximum. Instead, we use the AM-GM inequality to find the maximum value.

4. Apply the AM-GM inequality to \( x \):
   \[
   \frac{x}{27} = \left(\frac{a}{3}\right)^3 (1 - a) \leq \left(\frac{3 \cdot \frac{a}{3} + 1 - a}{4}\right)^4 = \frac{1}{256}
   \]
   Therefore, the maximum value of \( x \) is:
   \[
   A = \frac{27}{256}
   \]

5. Similarly, express \( y \) in terms of \( a \):
   \[
   y = \frac{a^2}{a^3 + 1}
   \]
   Apply the AM-GM inequality to \( y \):
   \[
   \frac{y}{4} = \left(\frac{a}{2}\right)^2 (1 - a) \leq \left(\frac{2 \cdot \frac{a}{2} + 1 - a}{3}\right)^3 = \frac{1}{27}
   \]
   Therefore, the maximum value of \( y \) is:
   \[
   B = \frac{4}{27}
   \]

6. Finally, calculate the ratio \( \frac{A}{B} \):
   \[
   \frac{A}{B} = \frac{\frac{27}{256}}{\frac{4}{27}} = \frac{27}{256} \cdot \frac{27}{4} = \frac{729}{1024}
   \]

The final answer is \( \boxed{\frac{729}{1024}} \).