a) Consider\[f\text{:}\left[ \text{0,}\infty  \right)\to \left[ \text{0,}\infty  \right)\] a differentiable and convex function .Show that $f\left( x \right)\le x$,  for every $x\ge 0$, than ${f}'\left( x \right)\le 1$ ,for every $x\ge 0$
b) Determine \[f\text{:}\left[ \text{0,}\infty  \right)\to \left[ \text{0,}\infty  \right)\] differentiable and convex functions which have the property that $f\left( 0 \right)=0\,$, and ${f}'\left( x \right)f\left( f\left( x \right) \right)=x$, for every $x\ge 0$
### Part (a)
1. **Assume for contradiction**: Suppose there exists some \( a \geq 0 \) such that \( f'(a) > 1 \).
2. **Convexity implies**: Since \( f \) is convex, \( f' \) is non-decreasing. Therefore, for all \( x \in [a, \infty) \), we have \( f'(x) \geq f'(a) \).
3. **Define a new function**: Let \( g(x) = f(x) - x \). Then, \( g'(x) = f'(x) - 1 \).
4. **Inequality for \( g'(x) \)**: For \( x \in [a, \infty) \), we have \( g'(x) \geq f'(a) - 1 \).
5. **Apply the Mean Value Theorem**: By the Mean Value Theorem, for any \( x > a \), there exists some \( t \in (a, x) \) such that
   \[
   \frac{g(x) - g(a)}{x - a} = g'(t).
   \]
6. **Substitute and simplify**:
   \[
   g(x) - g(a) = (x - a) g'(t) \geq (x - a) (f'(a) - 1).
   \]
   Therefore,
   \[
   g(x) \geq g(a) + (x - a) (f'(a) - 1).
   \]
7. **Contradiction**: Since \( f'(a) - 1 > 0 \), for sufficiently large \( x \), \( g(x) > 0 \). This implies \( f(x) > x \), which contradicts the given condition \( f(x) \leq x \) for all \( x \geq 0 \).

Thus, our assumption is false, and we conclude that \( f'(x) \leq 1 \) for all \( x \geq 0 \).

### Part (b)
1. **Given conditions**: \( f(0) = 0 \) and \( f'(x) f(f(x)) = x \) for all \( x \geq 0 \).
2. **Monotonicity**: Since \( f \) is differentiable and convex, \( f'(x) \geq 0 \). Also, \( f'(x) > 0 \) for \( x > 0 \), implying \( f \) is strictly increasing.
3. **Convexity implies non-negative second derivative**: Since \( f \) is convex, \( f''(x) \geq 0 \).
4. **Differentiate the given functional equation**:
   \[
   f'(x) f(f(x)) = x.
   \]
   Differentiating both sides with respect to \( x \), we get:
   \[
   f''(x) f(f(x)) + f'(x)^2 f'(f(x)) = 1.
   \]
5. **Substitute \( f'(x) = \frac{x}{f(f(x))} \)**:
   \[
   f''(x) f(f(x)) + \left( \frac{x}{f(f(x))} \right)^2 f'(f(x)) = 1.
   \]
6. **Simplify the expression**:
   \[
   f''(x) f(f(x)) + \frac{x^2 f'(f(x))}{f(f(x))^2} = 1.
   \]
7. **Substitute \( f'(f(x)) = \frac{f(x)}{f(f(f(x)))} \)**:
   \[
   f''(x) f(f(x)) + \frac{x^2 f(x)}{f(f(x))^2 f(f(f(x)))} = 1.
   \]
8. **Inequality analysis**: Since \( f''(x) \geq 0 \), we have:
   \[
   \frac{x^2 f(x)}{f(f(x))^2 f(f(f(x)))} \leq 1.
   \]
   This implies:
   \[
   f(f(x))^2 f(f(f(x))) \geq x^2 f(x).
   \]
9. **Assume \( f(a) < a \)**: If there exists \( a \) such that \( f(a) < a \), then \( f(f(a)) < a^2 \) and \( f(f(f(a))) < f(a) \), which contradicts the above inequality. Hence, \( f(x) \geq x \) for all \( x \geq 0 \).
10. **Combine inequalities**: Since \( f(x) \leq x \) and \( f(x) \geq x \), we conclude \( f(x) = x \) for all \( x \geq 0 \).

The final answer is \( \boxed{ f(x) = x } \).