If $\mathrm{f}(x)=3 \mathrm{x}+10, \mathrm{~g}(x)=x^2-1$, then $(\mathrm{fog})^{-1}(x)=$

To determine $(f \circ g)^{-1}(x)$, the composite function $(f \circ g)(x)$ must first be found, then inverted.

Given $f(x) = 3x + 10$ and $g(x) = x^2 - 1$, the composition is:
$(f \circ g)(x) = f(g(x)) = 3(x^2 - 1) + 10 = 3x^2 + 7$

Let $y = (f \circ g)(x) = 3x^2 + 7$. Interchanging $x$ and $y$ and solving for $y$:
$x = 3y^2 + 7 \implies y^2 = \frac{x - 7}{3} \implies y = \pm\sqrt{\frac{x - 7}{3}}$

Since the options only include the principal root, we take:
$(f \circ g)^{-1}(x) = \left(\frac{x - 7}{3}\right)^{\frac{1}{2}}$

This corresponds to option B.