Find $f$ such that $f^{\prime}(x)=x^{2}-6$ and $f(0)=1$.

Step 1 ：We are given the derivative of the function $f(x)$, which is $f'(x) = x^2 - 6$. We also know that $f(0) = 1$.

Step 2 ：To find the function $f(x)$, we need to integrate the derivative $f'(x)$ with respect to $x$. This will give us the antiderivative of $f'(x)$, which is $f(x)$.

Step 3 ：However, the integral will include a constant of integration, which we can determine using the condition $f(0) = 1$.

Step 4 ：The integral of $f'(x)$ is $\frac{x^{3}}{3} - 6x + C$, where $C$ is the constant of integration.

Step 5 ：Substituting $x = 0$ into the equation, we get $f(0) = C = 1$.

Step 6 ：Therefore, the function $f(x)$ that satisfies the given conditions is $f(x) = \frac{x^{3}}{3} - 6x + 1$.

Step 7 ：\(\boxed{f(x) = \frac{x^{3}}{3} - 6x + 1}\) is the final answer.