1. We need the number of real solutions of $f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0.$

2. Substitute $t=e^{2x}.$ Since $e^{2x}>0$ for all real $x$, we have $t>0$.

   Then the equation becomes $t^4-t^3-3t^2-t+1=0.$

3. Observe that the polynomial is palindromic: coefficients are $1,-1,-3,-1,1$. Divide by $t^2$ (valid since $t>0$): $t^2-t-3-\frac1t+\frac1{t^2}=0.$

   Rearranging, $\left(t^2+\frac1{t^2}\right)-\left(t+\frac1t\right)-3=0.$

4. Let $u=t+\frac1t.$ Then $t^2+\frac1{t^2}=u^2-2.$

   So the equation becomes $(u^2-2)-u-3=0,$ i.e. $u^2-u-5=0.$

5. Solve for $u$: $u=\frac{1\pm\sqrt{21}}{2}.$

6. Since $t>0$, we know $t+\frac1t\ge 2.$ Therefore only $u=\frac{1+\sqrt{21}}{2}$ is possible, because $\frac{1-\sqrt{21}}{2}<0$ is not allowed.

7. Now solve $t+\frac1t=\frac{1+\sqrt{21}}{2}.$ Multiplying by $t$: $t^2-\frac{1+\sqrt{21}}{2}t+1=0.$

   For the number of positive roots in $t$, check discriminant: $\Delta=\left(\frac{1+\sqrt{21}}{2}\right)^2-4.$

   Compute:

   =\frac{22+2\sqrt{21}}{4} =\frac{11+\sqrt{21}}{2}.$$ Hence $$\Delta=\frac{11+\sqrt{21}}{2}-4 =\frac{3+\sqrt{21}}{2}>0.$$ So there are two distinct real roots in $t$. Also their product is $1>0$ and sum is positive, hence both roots are positive.

8. Since $t=e^{2x}$ is a one-to-one correspondence between $x\in\mathbb R$ and $t>0$, each positive root of $t$ gives exactly one real $x$.

   Therefore, the curve cuts the $x$-axis at 2 points.

9. Comparison with stored answer: Stored correct answer = $2$. Our derived answer also = $2$.

Hence the number of points where the curve cuts the $x$-axis is $\boxed{2}.$