Find a function $f$ and a number $a$ such that \[ 1+\int_{a}^{x} \frac{f(t)}{t^{3}} d t=6 x^{-1} \] \[ f(x)= \] \[ a= \]

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 \(F'(x) = \frac{f(x)}{x^{3}}\) and \(F(x) = 6x^{-1} - 1\).

Step 4 ：We can find \(F'(x)\) 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 \(f(x) = \boxed{-6x}\) and \(a = \boxed{6}\).