1. Interpret the equation carefully

   The equation is

   $$\sin^{-1}\left(x^2+\frac13\right)+\cos^{-1}\left(x^2-\frac23\right)=x^2,$$

   for $x\in[-1,1]$.

   Here $[x]$ is mentioned in the statement, but it does not appear in the equation, so it is irrelevant.

2. Set a substitution

   Let

   $$t=x^2.$$

   Since $x\in[-1,1]$, we have

   $$t\in[0,1].$$

   The equation becomes

   $$\sin^{-1}\left(t+\frac13\right)+\cos^{-1}\left(t-\frac23\right)=t.$$

3. Apply domain restrictions of inverse trigonometric functions

   For $\sin^{-1}(u)$ to be defined, we need $u\in[-1,1]$. So,

   $$-1\le t+\frac13\le 1.$$

   Since $t\in[0,1]$, the left inequality is automatic, and the right gives

   $$t\le \frac23.$$

   For $\cos^{-1}(v)$ to be defined, we need $v\in[-1,1]$. So,

   $$-1\le t-\frac23\le 1.$$

   Since $t\in[0,1]$, this is always satisfied.

   Hence the common domain is

   $$t\in\left[0,\frac23\right].$$

4. Estimate the left-hand side

   For $t\in\left[0,\frac23\right]$:

   • $t+\frac13\in\left[\frac13,1\right]$, so

     $$\sin^{-1}\left(t+\frac13\right)\in\left[\sin^{-1}\left(\frac13\right),\frac\pi2\right].$$

     In particular,

     $$\sin^{-1}\left(t+\frac13\right)>0.$$

   • $t-\frac23\in\left[-\frac23,0\right]$, so

     $$\cos^{-1}\left(t-\frac23\right)\in\left[\cos^{-1}(0),\cos^{-1}\left(-\frac23\right)\right] =\left[\frac\pi2,\cos^{-1}\left(-\frac23\right)\right].$$

     Therefore,

     $$\cos^{-1}\left(t-\frac23\right)\ge \frac\pi2.$$

   So the entire left-hand side satisfies

   $$\sin^{-1}\left(t+\frac13\right)+\cos^{-1}\left(t-\frac23\right) > \frac\pi2.$$

5. Compare with the right-hand side

   But the right-hand side is $t$, and since $t\in\left[0,\frac23\right]$,

   $$t\le \frac23 < \frac\pi2.$$

   Hence,

   $$\sin^{-1}\left(t+\frac13\right)+\cos^{-1}\left(t-\frac23\right) > \frac\pi2 > t.$$

   Therefore the equation can never hold.

6. Conclusion

   There is no solution for $t$, hence no solution for $x\in[-1,1]$.

   So the number of solutions is

   $$\boxed{0}.$$

7. Option check

   • A: $0$ ✅
   • B: Infinite ❌
   • C: $2$ ❌
   • D: $4$ ❌

Therefore, the correct option is A.