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. Substitute $k=4$ and $k=6$: $f_4(x)=\frac{1}{4}(\sin^4 x+\cos^4 x),$ $f_6(x)=\frac{1}{6}(\sin^6 x+\cos^6 x).$ 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 $a=\sin^2 x, \qquad b=\cos^2 x.$ Then $a+b=1.$ Also, $\sin^4 x+\cos^4 x=a^2+b^2,$ $\sin^6 x+\cos^6 x=a^3+b^3.$ Hence, $f_4(x)-f_6(x)=\frac{1}{4}(a^2+b^2)-\frac{1}{6}(a^3+b^3).$

4. Use identities with $a+b=1$. First, $a^2+b^2=(a+b)^2-2ab=1-2ab.$ Next, $a^3+b^3=(a+b)^3-3ab(a+b)=1-3ab.$ Therefore, $f_4(x)-f_6(x)=\frac{1}{4}(1-2ab)-\frac{1}{6}(1-3ab).$

5. Simplify: $=\left(\frac{1}{4}-\frac{1}{6}\right)+\left(-\frac{2ab}{4}+\frac{3ab}{6}\right).$ Now, $\frac{1}{4}-\frac{1}{6}=\frac{1}{12},$ and $-\frac{2ab}{4}+\frac{3ab}{6}=-\frac{ab}{2}+\frac{ab}{2}=0.$ So, $f_4(x)-f_6(x)=\frac{1}{12}.$

6. Thus the value is constant and equals $\boxed{\frac{1}{12}}.$

7. Checking options:

• A: $\frac14$ ❌
• B: $\frac1{12}$ ✅
• C: $\frac16$ ❌
• D: $\frac13$ ❌

Therefore, the correct option is B.