Answer: \(\mathrm f^{-1}(x)=-\sqrt{\ln x-3}\), domain \(x\gt \mathrm e^3\), range \(\mathrm f^{-1}(x)\lt 0\), and \(\mathrm g(x)=-\sqrt{2x-3}\).

We start with the main method. Check whether the function is one-to-one on the stated domain before finding an inverse.

The function \(\mathrm f\) is defined by

\(\mathrm f(x)=\mathrm e^{x^2+3}\quad\text{for }x\lt 0.\)

On \(x\lt 0\), as \(x\) increases, \(x^2\) decreases. Hence \(\mathrm f\) is one-one, so \(\mathrm f^{-1}\) exists.

To find the inverse, let

\(y=\mathrm e^{x^2+3}.\)

Then

\(\ln y=x^2+3.\)

So

\(x^2=\ln y-3.\)

Since the original domain is \(x\lt 0\), take the negative square root:

\(x=-\sqrt{\ln y-3}.\)

Therefore

\(\mathrm f^{-1}(x)=-\sqrt{\ln x-3}.\)

The domain of \(\mathrm f^{-1}\) is the range of \(\mathrm f\). Since \(x\lt 0\), \(x^2\gt 0\), so \(x^2+3\gt 3\), and therefore

\(x\gt \mathrm e^3.\)

The range of \(\mathrm f^{-1}\) is the domain of \(\mathrm f\), so

\(\mathrm f^{-1}(x)\lt 0.\)

Now use \(\mathrm{fg}(x)=\mathrm e^{2x}\). This means

\(\mathrm f(\mathrm g(x))=\mathrm e^{2x}.\)

Apply \(\mathrm f^{-1}\) to both sides:

\(\mathrm g(x)=\mathrm f^{-1}(\mathrm e^{2x}).\)

Hence

\(\mathrm g(x)=-\sqrt{\ln(\mathrm e^{2x})-3}=-\sqrt{2x-3}.\)

This completes the solution and gives the required result.