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 $f^{\prime }(x)=x^2-6$. We also know that $f(0)=1$.

Step 2 ：To find the function $f(x)$, we need to integrate the derivative $f^{\prime }(x)$ with respect to $x$. This will give us the antiderivative of $f^{\prime }(x)$, 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 $f^{\prime }(x)$ 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 ：$\boxed{{\displaystyle f(x)=\frac{x^3}{3}-6x+1}}$ is the final answer.