Answer
inside: $g(x)= x^{2}-3 x$
outside: $f(g)=g^{1/2}$
derivative: $f^{\prime}(x)=\dfrac{ 2x-3}{2\sqrt{x^2-3x}}$
Work Step by Step
Given $$
f(x)=\sqrt{x^{2}-3 x}
$$
Rewriting $f(x) $ as$$
f(x)= (x^{2}-3 x)^{1/2}
$$
Use the chain rule to take the derivative
$$\frac{d f(g(x))}{d x}=f^{\prime}(g(x)) g^{\prime}(x)$$
Here $g(x)=x^{2}-3 x$ and $ f(g )= g^{1/2}$, then
\begin{align*}
f'(x)&=( g^{1/2}(x))'\\
&=\frac{1}{2}g^{-1/2}(x)g'(x)\\
&=\frac{1}{2}(x^2-3x)^{-1/2} (2x-3)\\
&=\frac{2x-3}{2\sqrt{x^2-3x}}
\end{align*}