First evaluate $I=\int_0^{\pi/2} \frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx.$

Use the standard property: If $J=\int_0^{\pi/2} \frac{f(\sin x,\cos x)}{f(\sin x,\cos x)+f(\cos x,\sin x)}\,dx,$ then by substituting $x\mapsto \frac{\pi}{2}-x$, we get $J=\int_0^{\pi/2} \frac{\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx.$ Adding both expressions, $2I=\int_0^{\pi/2}1\,dx=\frac{\pi}{2}.$ Hence, $I=\frac{\pi}{4}.$

Therefore the required integral becomes

Let $K=\int_0^{\pi/2} \frac{x\sin x\cos x}{\sin^4x+\cos^4x}\,dx.$ Apply the substitution $x\mapsto \frac{\pi}{2}-x$: $K=\int_0^{\pi/2} \frac{\left(\frac{\pi}{2}-x\right)\sin x\cos x}{\sin^4x+\cos^4x}\,dx,$ since under this substitution, $\sin x\cos x$ and $\sin^4x+\cos^4x$ remain unchanged.

Add the two forms of $K$: $2K=\int_0^{\pi/2} \frac{\left[x+\left(\frac{\pi}{2}-x\right)\right]\sin x\cos x}{\sin^4x+\cos^4x}\,dx$ $=\frac{\pi}{2}\int_0^{\pi/2} \frac{\sin x\cos x}{\sin^4x+\cos^4x}\,dx.$

So, $K=\frac{\pi}{4}\int_0^{\pi/2} \frac{\sin x\cos x}{\sin^4x+\cos^4x}\,dx.$

Now evaluate $L=\int_0^{\pi/2} \frac{\sin x\cos x}{\sin^4x+\cos^4x}\,dx.$ Put $t=\tan x$. Then $\sin x\cos x=\frac{t}{1+t^2},\qquad dx=\frac{dt}{1+t^2},$ and $\sin^4x+\cos^4x=\frac{t^4+1}{(1+t^2)^2}.$ Therefore,

Now let $u=t^2$, so $du=2t\,dt$:

Hence, $K=\frac{\pi}{4}\cdot \frac{\pi}{4}=\frac{\pi^2}{16}.$

So the required value is $\boxed{\frac{\pi^2}{16}}.$ This corresponds to Option C.

Comparison with stored answer: Stored correct answer is C, which matches our result.