1. Let $I=\int_{-\pi/2}^{\pi/2}\frac{dx}{(1+e^x)(\sin^6 x+\cos^6 x)}.$

We will use the standard symmetry property on $[-a,a]$: $\int_{-a}^{a} f(x)\,dx=\int_{-a}^{a} f(-x)\,dx,$ so that $2I=\int_{-\pi/2}^{\pi/2}\big(f(x)+f(-x)\big)dx,$ where $f(x)=\frac{1}{(1+e^x)(\sin^6 x+\cos^6 x)}.$

2. Compute $f(-x)$. Using $e^{-x}=\frac1{e^x},\qquad \sin(-x)=-\sin x,\qquad \cos(-x)=\cos x,$ and since powers are even, $\sin^6(-x)+\cos^6(-x)=\sin^6 x+\cos^6 x.$ Therefore,

Now, $\frac{1}{1+e^{-x}}=\frac{e^x}{1+e^x}.$ So, $f(-x)=\frac{e^x}{(1+e^x)(\sin^6 x+\cos^6 x)}.$

3. Add $f(x)$ and $f(-x)$: $f(x)+f(-x)=\frac{1}{(1+e^x)(\sin^6 x+\cos^6 x)}+\frac{e^x}{(1+e^x)(\sin^6 x+\cos^6 x)}.$ Hence, $f(x)+f(-x)=\frac{1+e^x}{(1+e^x)(\sin^6 x+\cos^6 x)}=\frac{1}{\sin^6 x+\cos^6 x}.$ Thus, $2I=\int_{-\pi/2}^{\pi/2}\frac{dx}{\sin^6 x+\cos^6 x}.$ So, $I=\frac12\int_{-\pi/2}^{\pi/2}\frac{dx}{\sin^6 x+\cos^6 x}.$

4. Simplify $\sin^6 x+\cos^6 x$. Use $a^3+b^3=(a+b)^3-3ab(a+b).$ Taking $a=\sin^2 x$, $b=\cos^2 x$, $\sin^6 x+\cos^6 x=(\sin^2 x)^3+(\cos^2 x)^3.$ Since $\sin^2 x+\cos^2 x=1,$ we get $\sin^6 x+\cos^6 x=1-3\sin^2 x\cos^2 x.$ Also, $\sin^2 x\cos^2 x=\frac14\sin^2 2x.$ Therefore,

So, $2I=\int_{-\pi/2}^{\pi/2}\frac{dx}{1-\frac34\sin^2 2x}.$

5. Use symmetry again. The integrand is even, so $2I=2\int_0^{\pi/2}\frac{dx}{1-\frac34\sin^2 2x}.$ Hence, $I=\int_0^{\pi/2}\frac{dx}{1-\frac34\sin^2 2x}.$ Now substitute $t=2x\quad\Rightarrow\quad dt=2dx,\quad dx=\frac{dt}{2}.$ As $x:0\to \pi/2$, we have $t:0\to \pi$. Thus, $I=\frac12\int_0^{\pi}\frac{dt}{1-\frac34\sin^2 t}.$ Using symmetry about $\pi/2$, $\int_0^{\pi}\frac{dt}{1-\frac34\sin^2 t}=2\int_0^{\pi/2}\frac{dt}{1-\frac34\sin^2 t}.$ Therefore, $I=\int_0^{\pi/2}\frac{dt}{1-\frac34\sin^2 t}.$

6. Use the standard result $\int_0^{\pi/2}\frac{d\theta}{1-m\sin^2\theta}=\frac{\pi}{2\sqrt{1-m}},\qquad (m<1).$ Here $m=\frac34$. So, $I=\frac{\pi}{2\sqrt{1-3/4}}=\frac{\pi}{2\sqrt{1/4}}=\frac{\pi}{2\cdot 1/2}=\pi.$

7. Compare with options:

• A: $2\pi$
• B: $0$
• C: $\pi$
• D: $\frac{\pi}{2}$

Hence the correct option is $\boxed{\text{C }(\pi)}.$