Let $f(x)=ax^3+bx^2+cx+d$ with $a\neq 0$ since the degree is $3$.

We are given:

1. $f(-1)=10$
2. $f(1)=-6$
3. $f(x)$ has a critical point at $x=-1 \Rightarrow f'(-1)=0$
4. $f'(x)$ has a critical point at $x=1 \Rightarrow f''(1)=0$

We will use these to determine $f(x)$ and then find where the local minimum occurs.

1. Compute derivatives

$f'(x)=3ax^2+2bx+c$ $f''(x)=6ax+2b$

Since $f'(x)$ has a critical point at $x=1$, $f''(1)=0$ So, $6a+2b=0 \Rightarrow b=-3a$

2. Use the condition $f'(-1)=0$

$f'(-1)=3a(-1)^2+2b(-1)+c=0$ $3a-2b+c=0$ Substitute $b=-3a$: $3a-2(-3a)+c=0$ $3a+6a+c=0$ $9a+c=0 \Rightarrow c=-9a$

3. Use the value conditions

From $f(1)=-6$:

$a+b+c+d=-6$ Substitute $b=-3a$, $c=-9a$: $a-3a-9a+d=-6$ $-11a+d=-6 \Rightarrow d=-6+11a$

From $f(-1)=10$:

$-a+b-c+d=10$ Substitute $b=-3a$, $c=-9a$: $-a-3a-(-9a)+d=10$ $-4a+9a+d=10$ $5a+d=10$ Now substitute $d=-6+11a$: $5a+(-6+11a)=10$ $16a-6=10$ $16a=16$ $a=1$

Hence, $b=-3,\, c=-9,\, d=5$

So, $f(x)=x^3-3x^2-9x+5$

4. Find critical points of $f(x)$

Differentiate: $f'(x)=3x^2-6x-9=3(x^2-2x-3)=3(x-3)(x+1)$ Thus the critical points are at $x=-1 \quad \text{and} \quad x=3$

5. Determine where the local minimum occurs

Use the second derivative: $f''(x)=6x-6$ At $x=-1$, $f''(-1)=6(-1)-6=-12<0$ So $x=-1$ is a local maximum.

At $x=3$, $f''(3)=18-6=12>0$ So $x=3$ is a local minimum.

Final Answer

$\boxed{3}$

The polynomial has a local minimum at $x=3$.