1. Use the condition on local extrema

Since $f(x)$ is a non-zero polynomial of degree $4$ and has local extreme points at $x=-1,0,1$, we must have

$f'(-1)=f'(0)=f'(1)=0.$

Now $f'(x)$ is a polynomial of degree $3$, so these three roots determine it completely up to a non-zero constant:

$f'(x)=k(x+1)x(x-1)=k(x^3-x), \qquad k\ne 0.$

2. Integrate to get $f(x)$

Integrating,

$f(x)=k\left(\frac{x^4}{4}-\frac{x^2}{2}\right)+C.$

So,

$f(0)=C.$

3. Solve $f(x)=f(0)$

We need the set

$S=\{x\in\mathbb R: f(x)=f(0)\}.$

Substitute:

$k\left(\frac{x^4}{4}-\frac{x^2}{2}\right)+C=C.$

Since $k\ne 0$,

$\frac{x^4}{4}-\frac{x^2}{2}=0.$

Multiply by $4$:

$x^4-2x^2=0.$

Factor:

$x^2(x^2-2)=0.$

Hence,

$x=0,\quad x=\pm\sqrt{2}.$

So,

$S=\{ -\sqrt{2},\ 0,\ \sqrt{2} \}.$

4. Classify the numbers

• $0$ is rational.
• $\pm\sqrt{2}$ are irrational.

Thus $S$ contains two irrational and one rational number.

5. Check options

• A: four rational numbers — false
• B: four irrational numbers — false
• C: two irrational and one rational number — true
• D: two irrational and two rational numbers — false

Therefore, the correct option is C.