Answer
$a=-1,b=3, c=9$
Work Step by Step
Step 1. Given $y=ax^3+bx^2+cx$, we have $y'=3ax^2+2bx+c$ and $y''=6ax+2b$
Step 2. The inflection point requires $y''=0$ which gives $x=-\frac{b}{3a}=1$ or $b=-3a$
Step 3. The extrema happen when $y'=0$ or $3ax^2+2bx+c=0$, which can be rewritten as $3ax^2-6ax+c=0$
Step 4. Since $x=-1,3$ are solutions to the above quadratic equation, we have $\begin{cases} 3a+6a+c=0 \\ 27a-18a+c=0 \end{cases}$ which gives $c=-9a$
Step 5. The original function becomes $y=ax^3-3ax^2-9ax$. As point $(1,11)$ is on the curve, we have $11=a-3a-9a$ or $a=-1$ which gives $b=3, c=9$.
Step 6. We can check the function $y=-x^3+3x^2+9x$ to have a local maximum at $x=3$ and a local minimum at $x=-1$ by the sign changes in $y'=-3x^2+6x+9$ across $x=-1,3$ as: $..(-)..(-1)..(+)..(3)..(-)..$
