1. We are given $f(3x)-f(x)=x \,\text{ for all } x\in \mathbb{R},$ and $f$ is continuous.

2. We want to find $f(14)$, given that $f(8)=7.$

3. Use the functional equation at a convenient value of $x$. To connect $8$ and $14$, notice that if we take $x=\frac{8}{3},$ then $3x=8$.

   Substituting into $f(3x)-f(x)=x,$ we get $f(8)-f\left(\frac{8}{3}\right)=\frac{8}{3}.$

   Since $f(8)=7$, $7-f\left(\frac{8}{3}\right)=\frac{8}{3}.$

   Hence, $f\left(\frac{8}{3}\right)=7-\frac{8}{3}=\frac{21-8}{3}=\frac{13}{3}.$

4. Now take $x=\frac{14}{3},$ so that $3x=14$. Then $f(14)-f\left(\frac{14}{3}\right)=\frac{14}{3}.$

   But this alone does not directly use $f(8)=7$. So instead, let us derive the general form of $f$.

5. Solve the functional equation. We try a linear function: $f(x)=ax+b.$ Then $f(3x)-f(x)=(3ax+b)-(ax+b)=2ax.$ This must equal $x$ for all $x$, so $2a=1 \implies a=\frac12.$

   Thus one candidate is $f(x)=\frac{x}{2}+b.$

6. Verify using the given value $f(8)=7$: $f(8)=\frac{8}{2}+b=4+b=7,$ so $b=3.$

   Therefore, $f(x)=\frac{x}{2}+3.$

7. Now compute $f(14)$: $f(14)=\frac{14}{2}+3=7+3=10.$

8. Check uniqueness. Let $g(x)=f(x)-\left(\frac{x}{2}+3\right).$ Then $g$ is continuous and $g(3x)-g(x)=0 \implies g(3x)=g(x).$ Replacing $x$ by $x/3^n$, $g(x)=g\left(\frac{x}{3^n}\right).$ As $n\to\infty$, $\frac{x}{3^n}\to 0$. By continuity, $g(x)=g(0),$ so $g$ is constant. Hence the linear form found is the unique continuous solution.

Therefore, $\boxed{f(14)=10}.$

So the correct option is B.