Answer
$a=2, b=-3, c=0, d=0$
Work Step by Step
Step 1. Let $f(x)=ax^3+bx^2+cx+d$; we have $f'(x)=3ax^2+2bx+c$
Step 2. As points $(0,0), (1,-1)$ are on the curve, we have $d=0$ and $a+b+c=-1$
Step 3. For point $(0,0)$ to be a local minimum of the function which does not have endpoints, we have $f'(0)=0$, which gives $c=0$
Step 4. For point $(1,-1)$ to be a local minimum of the function which does not have endpoints, we have $f'(1)=0$ which gives $3a+2b=0$
Step 5. Combine the above results to get $\begin{cases} a+b=-1\\3a+2b=0\end{cases}$.
Multiply $-2$ to the first equation and add the result to the second one to get $a=2$, which gives $b=-3$
Step 6. We conclude the results as $a=2, b=-3, c=0, d=0$
