 Determine twice differentiable functions $f: \mathbb{R} \rightarrow \mathbb{R}$ which verify relation 

    \[
    \left( f'(x) \right)^2 + f''(x) \leq 0, \forall x \in \mathbb{R}.
    \]
To determine the twice differentiable functions \( f: \mathbb{R} \rightarrow \mathbb{R} \) that satisfy the inequality 

\[
\left( f'(x) \right)^2 + f''(x) \leq 0, \forall x \in \mathbb{R},
\]

we will analyze the problem step by step.

1. **Introduce a new function \( g(x) \):**
   Let \( g(x) = f'(x) \). Then the given inequality can be rewritten in terms of \( g \) as:
   \[
   g(x)^2 + g'(x) \leq 0.
   \]

2. **Analyze the inequality \( g(x)^2 + g'(x) \leq 0 \):**
   This implies that:
   \[
   g'(x) \leq -g(x)^2.
   \]
   Since \( g'(x) \leq 0 \), \( g(x) \) is a non-increasing function.

3. **Consider the behavior of \( g(x) \):**
   - **Case 1: \( g(x) = 0 \) for all \( x \in \mathbb{R} \):**
     If \( g(x) = 0 \), then \( f'(x) = 0 \). Integrating this, we get:
     \[
     f(x) = C,
     \]
     where \( C \) is a constant.

   - **Case 2: \( g(x) \neq 0 \) for some \( x \in \mathbb{R} \):**
     Assume there exists some \( a \in \mathbb{R} \) such that \( g(a) \neq 0 \). Since \( g(x) \) is non-increasing, we need to consider the following sub-cases:
     
     - **Sub-case 2.1: \( g(a) > 0 \):**
       Since \( g(x) \) is non-increasing, \( g(x) \geq g(a) > 0 \) for \( x \leq a \). For \( x \leq a \), we have:
       \[
       \left( \frac{1}{g(x)} \right)' = -\frac{g'(x)}{g(x)^2} \geq 1.
       \]
       Integrating both sides from \( x \) to \( a \):
       \[
       \int_x^a \left( \frac{1}{g(t)} \right)' dt \geq \int_x^a 1 \, dt,
       \]
       which simplifies to:
       \[
       \frac{1}{g(a)} - \frac{1}{g(x)} \geq a - x.
       \]
       As \( x \to -\infty \), the left-hand side tends to \( -\infty \), which is a contradiction because \( g(x) \) would become negative for sufficiently large negative \( x \).

     - **Sub-case 2.2: \( g(a) < 0 \):**
       Since \( g(x) \) is non-increasing, \( g(x) \leq g(a) < 0 \) for \( x \geq a \). For \( x \geq a \), we have:
       \[
       \left( \frac{1}{g(x)} \right)' = -\frac{g'(x)}{g(x)^2} \geq 1.
       \]
       Integrating both sides from \( a \) to \( x \):
       \[
       \int_a^x \left( \frac{1}{g(t)} \right)' dt \geq \int_a^x 1 \, dt,
       \]
       which simplifies to:
       \[
       \frac{1}{g(x)} - \frac{1}{g(a)} \geq x - a.
       \]
       As \( x \to \infty \), the left-hand side tends to \( \infty \), which is a contradiction because \( g(x) \) would become positive for sufficiently large positive \( x \).

4. **Conclusion:**
   Since both sub-cases lead to contradictions, the only possibility is that \( g(x) = 0 \) for all \( x \in \mathbb{R} \). Therefore, \( f'(x) = 0 \) for all \( x \in \mathbb{R} \), which implies that \( f(x) \) is a constant function.

\(\blacksquare\)

The final answer is \( \boxed{ f(x) = C } \), where \( C \) is a constant.