Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ from $C^2$ (id est, $f$ is twice differentiable and $f''$ is continuous.) such that for every real number $t$ we have $f(t)^2=f(t \sqrt{2})$.
1. **Initial Setup and Proposition**:
   We are given the functional equation \( f(t)^2 = f(t\sqrt{2}) \) for all \( t \in \mathbb{R} \). We denote this proposition as \( P(t) \).

2. **Analyzing \( P(t/\sqrt{2}) \)**:
   From \( P(t/\sqrt{2}) \), we have:
   \[
   f\left(\frac{t}{\sqrt{2}}\right)^2 = f(t)
   \]
   This implies that \( f(t) \geq 0 \) for all \( t \in \mathbb{R} \).

3. **Case 1: \( f(t) = 0 \) for some \( t \in \mathbb{R} \)**:
   If \( f(t) = 0 \) for some \( t \), then by continuity of \( f \), we have:
   \[
   f(0) = f\left(\lim_{n \to \infty} \frac{t}{(\sqrt{2})^n}\right) = \lim_{n \to \infty} f\left(\frac{t}{(\sqrt{2})^n}\right) = 0
   \]
   If there exists \( y \in \mathbb{R} \) such that \( f(y) > 0 \), then:
   \[
   0 = f(0) = f\left(\lim_{n \to \infty} \frac{y}{(\sqrt{2})^n}\right) = \lim_{n \to \infty} f\left(\frac{y}{(\sqrt{2})^n}\right) = \lim_{n \to \infty} f(y)^{\frac{1}{2^n}} = f(y)^0 = 1
   \]
   This is a contradiction. Therefore, if \( f(0) = 0 \), then \( f \equiv 0 \).

4. **Case 2: \( f(0) > 0 \)**:
   Assume \( f(0) = 1 \). This implies \( f(x) > 0 \) for all \( x \in \mathbb{R} \).

5. **Defining \( g(x) = \ln(f(x)) \)**:
   Since \( f(x) > 0 \), \( g(x) \) is well-defined and twice differentiable. Differentiating \( f \) and \( g \), we get:
   \[
   2f(t)f'(t) = \sqrt{2}f'(t\sqrt{2})
   \]
   At \( t = 0 \):
   \[
   f'(0) = 0
   \]
   Therefore:
   \[
   g'(t) = \frac{f'(t)}{f(t)} \implies g'(0) = \frac{f'(0)}{f(0)} = 0
   \]

6. **Defining \( h(x) = \frac{g(x)}{x^2} \) for \( x \neq 0 \)**:
   The limit of \( h \) at \( 0 \) exists by L'Hôpital's rule:
   \[
   \lim_{x \to 0} h(x) = \lim_{x \to 0} \frac{g(x)}{x^2} = \lim_{x \to 0} \frac{g'(x)}{2x} = \lim_{x \to 0} \frac{g''(x)}{2} = \frac{g''(0)}{2} = L
   \]
   for some real number \( L \).

7. **Functional Equation for \( h \)**:
   From \( 2g(x) = g(x\sqrt{2}) \), we get:
   \[
   \frac{g(x)}{x^2} = \frac{g(x\sqrt{2})}{2x^2} \implies h(x) = h(x\sqrt{2})
   \]
   for all \( x \neq 0 \).

8. **Conclusion**:
   Since \( h \) has a limit at \( 0 \), for every \( a, b \in \mathbb{R} \), we can find large enough powers of \( \sqrt{2} \) such that their distance to \( 0 \) is close. Thus:
   \[
   |h(a) - h(b)| \leq |h(a) - L| + |h(b) - L| < 2\epsilon
   \]
   Taking \( \epsilon \to 0 \), we have \( h(a) = h(b) \) for all \( a, b \in \mathbb{R} \). Therefore, \( h(x) = c \) and:
   \[
   g(x) = cx^2 \implies f(x) = e^{cx^2}
   \]
   This satisfies the original equation.

The final answer is \( \boxed{ f(x) = e^{cx^2} } \).