Find maximum of the expression $(a -b^2)(b - a^2)$, where $0 \le  a,b  \le 1$.
To find the maximum of the expression \((a - b^2)(b - a^2)\) where \(0 \le a, b \le 1\), we can proceed as follows:

1. **Consider the expression \((a - b^2)(b - a^2)\):**
   \[
   f(a, b) = (a - b^2)(b - a^2)
   \]

2. **Analyze the given constraints:**
   \[
   0 \le a \le 1 \quad \text{and} \quad 0 \le b \le 1
   \]

3. **Use the method of completing the square to find a bound:**
   Consider the sum of squares:
   \[
   \left( a - \frac{1}{2} \right)^2 + \left( b - \frac{1}{2} \right)^2 \geq 0
   \]
   Expanding and simplifying, we get:
   \[
   a^2 - a + \frac{1}{4} + b^2 - b + \frac{1}{4} \geq 0
   \]
   \[
   a^2 + b^2 - a - b + \frac{1}{2} \geq 0
   \]
   Rearranging, we have:
   \[
   a + b - a^2 - b^2 \leq \frac{1}{2}
   \]

4. **Square the inequality to find a bound for the product:**
   \[
   \left( a + b - a^2 - b^2 \right)^2 \leq \left( \frac{1}{2} \right)^2
   \]
   \[
   \left( a + b - a^2 - b^2 \right)^2 \leq \frac{1}{4}
   \]

5. **Relate this to the original expression:**
   Notice that:
   \[
   (a - b^2)(b - a^2) \leq \frac{(a + b - a^2 - b^2)^2}{4}
   \]
   Therefore:
   \[
   (a - b^2)(b - a^2) \leq \frac{1}{16}
   \]

6. **Check for equality:**
   Equality holds when:
   \[
   a = b = \frac{1}{2}
   \]
   Substituting \(a = b = \frac{1}{2}\) into the original expression:
   \[
   \left( \frac{1}{2} - \left( \frac{1}{2} \right)^2 \right) \left( \frac{1}{2} - \left( \frac{1}{2} \right)^2 \right) = \left( \frac{1}{2} - \frac{1}{4} \right) \left( \frac{1}{2} - \frac{1}{4} \right) = \left( \frac{1}{4} \right) \left( \frac{1}{4} \right) = \frac{1}{16}
   \]

Thus, the maximum value of the expression \((a - b^2)(b - a^2)\) is \(\frac{1}{16}\).

The final answer is \(\boxed{\frac{1}{16}}\).