If $\mathrm{f}(x)=\log \left(\frac{1+x}{1-x}\right)$ and $\mathrm{g}(x)=\frac{3 x+x^3}{1+3 x^2}$, then $($ fog $)(x)=$

To compute the composition $(f \circ g)(x) = f(g(x))$, substitute $g(x)$ into $f$:

$f(g(x)) = \log\left(\frac{1 + g(x)}{1 - g(x)}\right)$

Given $g(x) = \frac{3x + x^3}{1 + 3x^2}$, we find:

$1 + g(x) = \frac{(1+x)^3}{1+3x^2}$,
$1 - g(x) = \frac{(1-x)^3}{1+3x^2}$

Substituting these into the expression:

$f(g(x)) = \log\left(\frac{(1+x)^3/(1+3x^2)}{(1-x)^3/(1+3x^2)}\right) = \log\left(\frac{(1+x)^3}{(1-x)^3}\right) = \log\left(\left(\frac{1+x}{1-x}\right)^3\right)$

Apply the logarithmic identity:

$f(g(x)) = 3\log\left(\frac{1+x}{1-x}\right) = 3f(x)$

This confirms option $\boxed{B}$ is correct.