1. We are given $f_k(x)=\frac{1}{k}\left(\sin^k x+\cos^k x\right).$

We need to find: $f_4(x)-f_6(x).$

2. Write both terms explicitly: $f_4(x)=\frac{1}{4}(\sin^4 x+\cos^4 x),$ $f_6(x)=\frac{1}{6}(\sin^6 x+\cos^6 x).$

So, $f_4(x)-f_6(x)=\frac{1}{4}(\sin^4 x+\cos^4 x)-\frac{1}{6}(\sin^6 x+\cos^6 x).$

3. Let $a=\sin^2 x,\quad b=\cos^2 x.$ Then $a+b=1.$ Also, $\sin^4 x+\cos^4 x=a^2+b^2,$ $\sin^6 x+\cos^6 x=a^3+b^3.$

Thus, $f_4(x)-f_6(x)=\frac{1}{4}(a^2+b^2)-\frac{1}{6}(a^3+b^3).$

4. Use identities with $a+b=1$.

First, $a^2+b^2=(a+b)^2-2ab=1-2ab.$

Second, $a^3+b^3=(a+b)^3-3ab(a+b)=1-3ab.$

Therefore, $f_4(x)-f_6(x)=\frac{1}{4}(1-2ab)-\frac{1}{6}(1-3ab).$

5. Expand: $=\frac{1}{4}-\frac{ab}{1?}??$ Let us do it carefully: $\frac{1}{4}(1-2ab)=\frac{1}{4}-\frac{1}{2}ab,$ $\frac{1}{6}(1-3ab)=\frac{1}{6}-\frac{1}{2}ab.$

Hence, $f_4(x)-f_6(x)=\left(\frac{1}{4}-\frac{1}{2}ab\right)-\left(\frac{1}{6}-\frac{1}{2}ab\right).$

The $ab$ terms cancel: $f_4(x)-f_6(x)=\frac{1}{4}-\frac{1}{6}=\frac{3-2}{12}=\frac{1}{12}.$

6. This is independent of $x$, so for all $x\in\mathbb{R}$, $f_4(x)-f_6(x)=\frac{1}{12}.$

7. Comparing with the options:

• A: $\frac14$
• B: $\frac{5}{12}$
• C: $-\frac{1}{12}$
• D: $\frac{1}{12}$

So the correct option is D.