1. Interpret the loci

We need the intersection of:

• $|z|=3$
• $\arg(z-1)-\arg(z+1)=\dfrac{\pi}{4}$

Let $z=x+iy.$ Then the first locus is the circle $x^2+y^2=9.$

2. Interpret the argument condition geometrically

The quantity $\arg(z-1)-\arg(z+1)$ is the angle between the vectors from the origin of the complex numbers $z+1$ and $z-1$. Equivalently, it is the angle subtended by the line segment joining $-1$ and $1$ at the point $z$.

So if $A(1,0)$ and $B(-1,0)$, then $\arg(z-1)-\arg(z+1)=\angle AZB.$ Thus the condition means: $\angle AZB=\frac{\pi}{4}.$

The set of points from which the chord $AB$ subtends a constant angle $\pi/4$ is a pair of circles through $A$ and $B$.

3. Find those circles explicitly

Here $AB=2.$ If a chord of a circle subtends angle $\theta$ at a point on the circle, then for radius $R$, $AB=2R\sin\theta.$ With $\theta=\pi/4$, $2=2R\sin\frac{\pi}{4}=2R\cdot \frac{1}{\sqrt2}.$ So $R=\sqrt2.$

The chord $AB$ has midpoint $(0,0)$, and the center lies on the perpendicular bisector, i.e. the $y$-axis. If the center is $(0,k)$, then $\sqrt{1+k^2}=R=\sqrt2,$ so $k^2=1 \implies k=\pm 1.$

Hence the two circles are $x^2+(y-1)^2=2$ and $x^2+(y+1)^2=2.$

4. Intersect with $x^2+y^2=9$

First with $x^2+(y-1)^2=2.$ Expanding, $x^2+y^2-2y+1=2$ $x^2+y^2-2y=1.$ Using $x^2+y^2=9$, $9-2y=1 \implies 2y=8 \implies y=4.$ But on the circle $x^2+y^2=9$, $|y|\le 3$, so impossible.

Now with $x^2+(y+1)^2=2.$ Expanding, $x^2+y^2+2y+1=2$ $x^2+y^2+2y=1.$ Using $x^2+y^2=9$, $9+2y=1 \implies 2y=-8 \implies y=-4,$ again impossible.

So the circle $|z|=3$ does not meet either of these loci.

5. Algebraic check using tangent formula

For completeness, $\arg(z-1)-\arg(z+1)=\arg\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}.$ So $\frac{z-1}{z+1}$ has argument $\pi/4$, meaning its imaginary and real parts are equal and positive.

Let $z=x+iy$. Then

Multiplying numerator and denominator by $x+1-iy$, \frac{z-1}{z+1}=rac{(x^2+y^2-1)+2iy}{(x+1)^2+y^2}. Thus $\tan\left(\arg(z-1)-\arg(z+1)\right)=\frac{2y}{x^2+y^2-1}.$ Since the angle is $\pi/4$, $\frac{2y}{x^2+y^2-1}=1.$ Using $x^2+y^2=9$, $\frac{2y}{8}=1 \implies y=4,$ which is impossible on $x^2+y^2=9$. So again, no intersection.

6. Conclusion

The two loci do not intersect at any point. Therefore the correct option is: $\boxed{\text{C: nowhere}}$