First, this has nothing to do with implicit differentiation; it’s just an ordinary derivative. Secondly, $\sec^{-1}x$ is most definitely not $\frac1{x\sqrt{x^2-1}}$; what you mean is that the derivative of $\sec^{-1}x$ with respect to $x$ is $\frac1{x\sqrt{x^2-1}}$.
Now let’s look at the actual differentiation problem. You have $f(x)=\sec^{-1}\sqrt{3x}$. Differentiating this will require the use of the chain rule, but not the product rule:
$\begin{align*} \left[\sec^{-1}\sqrt{3x}\right]'&=\frac1{\sqrt{3x}\sqrt{\left(\sqrt{3x}\right)^2-1}}\cdot\left[\sqrt{3x}\right]'\\ &=\frac1{\sqrt{3x}\sqrt{3x-1}}\cdot\left[(3x)^{1/2}\right]'\\ &=\frac1{\sqrt{3x}\sqrt{3x-1}}\cdot\frac12(3x)^{-1/2}\cdot[3x]'\\ &=\frac1{\sqrt{3x}\sqrt{3x-1}}\cdot\frac12\cdot\frac1{\sqrt{3x}}\cdot3\\ &=\frac3{2\sqrt{3x}\sqrt{3x}\sqrt{3x-1}}\\ &=\frac3{6x\sqrt{3x-1}}\\ &=\frac1{2x\sqrt{3x-1}} \end{align*}$
You’re never differentiating a product, so you never use the product rule. (Well, you could use it to differentiate $3x$, but that would be making a great deal of unnecessary work for yourself.)