Answer
$-\dfrac{1}{2r^2l} \sqrt{\dfrac{T}{\pi \omega}}, -\dfrac{1}{2rl^2} \sqrt{\dfrac{T}{\pi \omega}}, \dfrac{1}{4rl\sqrt{T\pi \omega}}, -\dfrac{\sqrt T}{4rl \omega\sqrt{\pi \omega}}$
Work Step by Step
$\dfrac{\partial f}{\partial r}=-\dfrac{1}{2r^2l} \sqrt{\dfrac{T}{\pi \omega}}$
and
$\dfrac{\partial f}{\partial l}=-\dfrac{1}{2rl^2} \sqrt{\dfrac{T}{\pi \omega}}$
and
$\dfrac{\partial f}{\partial T}=\dfrac{1}{4rl\sqrt{T\pi \omega}}$
and $\dfrac{\partial f}{\partial \omega}=\dfrac{1}{2rl}\sqrt {\dfrac{T}{\pi}}(-1/2 \omega ^{-3/2})=-\dfrac{\sqrt T}{4rl \omega\sqrt{\pi \omega}}$
Hence, $-\dfrac{1}{2r^2l} \sqrt{\dfrac{T}{\pi \omega}}, -\dfrac{1}{2rl^2} \sqrt{\dfrac{T}{\pi \omega}}, \dfrac{1}{4rl\sqrt{T\pi \omega}}, -\dfrac{\sqrt T}{4rl \omega\sqrt{\pi \omega}}$