Why don't we use $\tan^{-1}\frac a h= \frac \pi 2-\cot^{-1}\frac a h=\frac \pi 2-\tan^{-1}\frac h a.$
So, $\frac{d(\tan^{-1}\frac a h)}{dh}=\frac{d(\frac \pi 2-\tan^{-1}\frac h a)}{dh}=-\frac{d(\tan^{-1}\frac h a)}{dh}=-\frac{1}{1+(\frac h a)^2}\frac 1 a=-\frac{a}{h^2+a^2}$
So, $\frac{d(\tan^{-1}\frac {40} h)}{dh}=-\frac{40}{h^2+40^2}$ and
$\frac{d(\tan^{-1}\frac {32} h)}{dh}=-\frac{32}{h^2+32^2}$
So, $\frac{dy}{dh}=-\frac{40}{h^2+40^2}-\left(-\frac{32}{h^2+32^2}\right)=\frac{32}{h^2+32^2}-\frac{40}{h^2+40^2}=\frac{32\cdot 40\cdot 8-8h^2}{(h^2+32^2)(h^2+40^2)}$