Given $\alpha=\cos^{-1}\left(\frac35\right),\qquad \beta=\tan^{-1}\left(\frac13\right),\qquad 0<\alpha,\beta<\frac\pi2$

Find trigonometric ratios for $\alpha$

Since $\cos\alpha=\frac35,$ and $\alpha$ is in the first quadrant, using a right triangle: $\sin\alpha=\frac45,\qquad \tan\alpha=\frac43.$

Find trigonometric ratios for $\beta$

Since $\tan\beta=\frac13,$ and $\beta$ is in the first quadrant, take opposite $=1$, adjacent $=3$, so hypotenuse $=\sqrt{10}$. Hence, $\sin\beta=\frac1{\sqrt{10}},\qquad \cos\beta=\frac3{\sqrt{10}}.$

Compute $\sin(\alpha-\beta)$

$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$ $=\frac45\cdot\frac3{\sqrt{10}}-\frac35\cdot\frac1{\sqrt{10}}$ $=\frac{12}{5\sqrt{10}}-\frac{3}{5\sqrt{10}}$ $=\frac{9}{5\sqrt{10}}.$

Determine the angle uniquely

Since $0<\beta<\alpha<\frac\pi2$, we have $0<\alpha-\beta<\frac\pi2.$ Therefore, $\alpha-\beta$ is the principal acute angle whose sine is $\frac{9}{5\sqrt{10}}$.

So, $\alpha-\beta=\sin^{-1}\left(\frac{9}{5\sqrt{10}}\right).$

Check other equivalent forms (optional)

Also, $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$ $=\frac35\cdot\frac3{\sqrt{10}}+\frac45\cdot\frac1{\sqrt{10}}$ $=\frac{13}{5\sqrt{10}},$ so option C is not correct.

And,

Final Answer

$\boxed{\sin^{-1}\left(\frac{9}{5\sqrt{10}}\right)}$

Hence, Option B is correct.