If $\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{3}{5}+\sin ^{-1} x=\frac{\pi}{2} \quad$ then $x=$

Given: $\sin^{-1} \frac{1}{3} + \sin^{-1} \frac{3}{5} + \sin^{-1} x = \frac{\pi}{2}$

Rewriting the equation: $\sin^{-1} \frac{1}{3} + \sin^{-1} \frac{3}{5} = \frac{\pi}{2} - \sin^{-1} x$

Using the identity $\frac{\pi}{2} - \sin^{-1} x = \cos^{-1} x$, we obtain:

$\sin^{-1} \frac{1}{3} + \sin^{-1} \frac{3}{5} = \cos^{-1} x$

Let $\alpha = \sin^{-1} \frac{1}{3}$ and $\beta = \sin^{-1} \frac{3}{5}$, so that $\sin \alpha = \frac{1}{3}$ and $\sin \beta = \frac{3}{5}$.

Since $\alpha, \beta \in [0, \frac{\pi}{2}]$, both angles are in the first quadrant and their cosines are positive:

$\cos \alpha = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \frac{2\sqrt{2}}{3}$

$\cos \beta = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$

The equation becomes $\alpha + \beta = \cos^{-1} x$, so taking cosine of both sides:

$\cos(\alpha + \beta) = x$

Applying the cosine addition formula:

$x = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \left(\frac{2\sqrt{2}}{3}\right)\left(\frac{4}{5}\right) - \left(\frac{1}{3}\right)\left(\frac{3}{5}\right)$

Simplifying: $x = \frac{8\sqrt{2}}{15} - \frac{3}{15} = \frac{8\sqrt{2} - 3}{15}$

This corresponds to option D.

Final answer: $\boxed{\text{D}}$