Find the greatest real number $k$ such that, for any positive $a,b,c$ with $a^{2}>bc$, $(a^{2}-bc)^{2}>k(b^{2}-ca)(c^{2}-ab)$.
1. **Assume \(abc = 1\):**
   Without loss of generality, we can assume that \(abc = 1\). This assumption simplifies the problem without loss of generality because we can always scale \(a\), \(b\), and \(c\) to satisfy this condition.

2. **Substitute \(a\), \(b\), and \(c\) with \(x\), \(y\), and \(z\):**
   Let \(x = a^3\), \(y = b^3\), and \(z = c^3\). Then, since \(abc = 1\), we have \(xyz = 1\). Also, since \(a^2 > bc\), we have \(x > 1\).

3. **Rewrite the inequality:**
   The given inequality \((a^2 - bc)^2 > k(b^2 - ca)(c^2 - ab)\) can be rewritten in terms of \(x\), \(y\), and \(z\) as:
   \[
   \left(\frac{x - 1}{x}\right)^2 > k \left(\frac{y - 1}{y}\right) \left(\frac{z - 1}{z}\right)
   \]

4. **Simplify the inequality:**
   Since \(xyz = 1\), we can rewrite the inequality as:
   \[
   \frac{(x - 1)^2}{x} > k (y - 1)(z - 1)
   \]

5. **Consider the case \((y - 1)(z - 1) \leq 0\):**
   If \((y - 1)(z - 1) \leq 0\), the inequality is trivially true because the left-hand side is positive and the right-hand side is non-positive.

6. **Consider the case \((y - 1)(z - 1) > 0\):**
   If \((y - 1)(z - 1) > 0\), then both \(y > 1\) and \(z > 1\). We need to find the minimum value of \(\frac{(x - 1)^2}{x(y - 1)(z - 1)}\).

7. **Simplify further:**
   Since \(xyz = 1\), we have \(z = \frac{1}{xy}\). Substitute \(z\) into the inequality:
   \[
   \frac{(x - 1)^2}{x(y - 1)\left(\frac{1}{xy} - 1\right)} = \frac{(x - 1)^2}{x(y - 1)\left(\frac{1 - xy}{xy}\right)} = \frac{(x - 1)^2}{x(y - 1)\left(\frac{1 - xy}{xy}\right)} = \frac{(x - 1)^2 \cdot xy}{x(y - 1)(1 - xy)}
   \]

8. **Simplify the expression:**
   \[
   \frac{(x - 1)^2 \cdot xy}{x(y - 1)(1 - xy)} = \frac{(x - 1)^2 \cdot y}{(y - 1)(1 - xy)}
   \]

9. **Find the minimum value:**
   To find the minimum value, we need to analyze the behavior of the function. Notice that as \(x \to 1\), the expression \(\frac{(x - 1)^2}{x}\) approaches 0. However, for \(x > 1\), the expression \(\frac{(x - 1)^2}{x}\) increases. The minimum value of \(\frac{(x - 1)^2}{x}\) for \(x > 1\) is achieved when \(x = 2\), giving us:
   \[
   \frac{(2 - 1)^2}{2} = \frac{1}{2}
   \]

10. **Conclusion:**
    Therefore, the greatest real number \(k\) such that the inequality holds for any positive \(a, b, c\) with \(a^2 > bc\) is \(k = 4\).

The final answer is \(\boxed{4}\).