If $f(x)=\int_{0}^{x}\left(t^{3}+6 t^{2}+3\right) d t$ then \[ f^{\prime \prime}(x)= \]

Step 1 ：The function $f(x)$ is defined as the integral of a function from 0 to $x$. The Fundamental Theorem of Calculus tells us that the derivative of an integral of a function from a constant to $x$ is just the function evaluated at $x$. Therefore, the first derivative of $f(x)$ is $f'(x) = x^3 + 6x^2 + 3$.

Step 2 ：To find the second derivative, we need to take the derivative of $f'(x)$.

Step 3 ：The second derivative of the function $f(x)$ is $f''(x) = 3x^2 + 12x$.

Step 4 ：Final Answer: \(\boxed{f''(x) = 3x^2 + 12x}\)