Step 1: Simplify the denominator.
$\sin^6x+\cos^6x=(\sin^2x)^3+(\cos^2x)^3$. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ with $a+b=1$, it becomes $\sin^4x-\sin^2x\cos^2x+\cos^4x$.
Step 2: Reduce further.
Since $\sin^4x+\cos^4x=1-2\sin^2x\cos^2x$, the denominator becomes $1-3\sin^2x\cos^2x$.
Step 3: Switch to $t=\tan x$.
Put $t=\tan x$, so $dx=\dfrac{dt}{1+t^2}$ and $\sin^2x\cos^2x=\dfrac{t^2}{(1+t^2)^2}$. After substituting and simplifying, the integral becomes \[ I=\int\frac{t^2}{1+t^6}\,dt. \]
Step 4: Substitute again.
Let $u=t^3$, then $du=3t^2\,dt$, so $t^2\,dt=\dfrac{du}{3}$.
Step 5: Integrate the simple form.
\[ I=\frac13\int\frac{du}{1+u^2}=\frac13\tan^{-1}u+c. \]
Step 6: Put everything back.
Since $u=t^3=\tan^3x$, \[ I=\frac13\tan^{-1}(\tan^3x)+c. \] \[ \boxed{\dfrac13\tan^{-1}(\tan^3x)+c} \]