Find a function $f$ and a number $a$ such that
$$1+{\int }_a^x\frac{f(t)}{t^3}dt=6x^{-1}$$
$$f(x)=$$
$$a=$$

Step 1 ：The given equation is in the form of the Fundamental Theorem of Calculus, which states that if a function f is continuous over the interval [a, b] and F is an antiderivative of f on [a, b], then ${\int }_a^{b}f(x)dx=F(b)-F(a)$.

Step 2 ：In this case, we have $1+{\int }_a^x\frac{f(t)}{t^3}dt=6x^{-1}$. We can rewrite the right side of the equation as $6x^{-1}-1$ to match the format of the Fundamental Theorem of Calculus.

Step 3 ：So, we need to find a function F(x) such that $F^{\prime }(x)=\frac{f(x)}{x^3}$ and $F(x)=6x^{-1}-1$.

Step 4 ：We can find $F^{\prime }(x)$ by differentiating $F(x)$, and then multiply the result by $x^3$ to find $f(x)$.

Step 5 ：To find a, we need to solve the equation $F(a)=0$.

Step 6 ：By solving the above steps, we find that the function $f(x)$ is $f(x)=-6x$ and the number a is $a=6$.

Step 7 ：So, we have $f(x)=\boxed{{\displaystyle -6x}}$ and $a=\boxed{{\displaystyle 6}}$.