Given that,
$$
f(x)=\frac{1}{x^2} \int_3^x\left(2 t-3 f^{\prime}(t)\right) d t
$$
On differentiating both sides w.r.t. $x$, we get
$$
\begin{aligned}
& f^{\prime}(x)=\frac{1}{x^2} \frac{d}{d x}\left(\int_3^x\left(2 t-3 f^{\prime}(t) d t\right)\right. \\
& \quad+\int_3^x\left(2 t-3 f^{\prime}(t)\right) d t \frac{d}{d x}\left(\frac{1}{x^2}\right)
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad f^{\prime}(x)=\frac{1}{x^2}\left(2 x-3 f^{\prime}(x)\right) \\
&-\frac{2}{x^3} \int_3^x\left(2 t-3 f^{\prime}(t)\right) d t
\end{aligned}
$$
At $x=3$,
$$
\begin{aligned}
f^{\prime}(3) & =\frac{1}{3^2}\left(6-3 f^{\prime}(3)\right)-0 \\
\Rightarrow \quad \frac{4 f^{\prime}(3)}{3} & =\frac{2}{3} \Rightarrow f^{\prime}(3)=\frac{1}{2}
\end{aligned}
$$