1. Equation of all lines parallel to $4x-3y+2=0$

Any line parallel to $4x-3y+2=0$ has the form $4x-3y+c=0$ for some constant $c$.

2. Use the distance from origin condition

Distance of the point $(0,0)$ from the line $4x-3y+c=0$ is $\frac{|c|}{\sqrt{4^2+(-3)^2}}=\frac{|c|}{5}$

Given distance is $\dfrac{3}{5}$, so $\frac{|c|}{5}=\frac{3}{5}$ $|c|=3$ Hence the required lines are $4x-3y+3=0$ and $4x-3y-3=0$

3. Check which option lies on any of these lines

A point lies on one of these lines if $4x-3y=\pm 3$

Option A: $\left(\dfrac14,-\dfrac13\right)$

$4\left(\frac14\right)-3\left(-\frac13\right)=1+1=2$ Not equal to $\pm 3$.

Option B: $\left(-\dfrac14,\dfrac23\right)$

$4\left(-\frac14\right)-3\left(\frac23\right)=-1-2=-3$ This satisfies $4x-3y=-3$, so it lies on $4x-3y+3=0$

Option C: $\left(-\dfrac14,-\dfrac23\right)$

$4\left(-\frac14\right)-3\left(-\frac23\right)=-1+2=1$ Not equal to $\pm 3$.

Option D: $\left(\dfrac14,\dfrac13\right)$

$4\left(\frac14\right)-3\left(\frac13\right)=1-1=0$ Not equal to $\pm 3$.

4. Conclusion

The point that lies on one of the required lines is $\boxed{\left(-\frac14,\frac23\right)}$ which is Option B.