If $f(x)=\int_{3}^{x} t^{8} d t$
then
\[
\begin{array}{l}
f^{\prime}(x)= \\
f^{\prime}(3)=
\end{array}
\]

Step 1 ：The function $f(x)$ is defined as an integral from a constant to $x$ of a function of $t$. The Fundamental Theorem of Calculus tells us that the derivative of such a function is simply the function that we're integrating, evaluated at $x$.

Step 2 ：So, to find $f'(x)$, we simply need to replace $t$ with $x$ in the function we're integrating, $t^8$. This gives us $f'(x) = x^8$.

Step 3 ：Then, to find $f'(3)$, we substitute $x=3$ into our derivative function. This gives us $f'(3) = 3^8 = 6561$.

Step 4 ：Final Answer: The derivative of the function $f(x)$ is $f'(x) = \boxed{x^8}$ and the value of the derivative at $x=3$ is $f'(3) = \boxed{6561}$.