Answer
$\frac{dr}{d\theta}=\frac{-1}{6}$
Work Step by Step
Given $r=(\theta^2+7)^{(\frac{1}{3})}$ and
$\theta^2 t+\theta=1$
On differentiating both sides:
$\frac{dr}{d\theta}=\frac{1}{3}(({\theta^2+7})^{\frac{-2}{3}}\frac{d({\theta^2+7})}{d\theta}$
$\frac{dr}{d\theta}=\frac{(2\theta)}{3}(({\theta^2+7})^{\frac{-2}{3}}\frac{d\theta}{dt}$
and $2\theta{t}\frac{d\theta}{dt}+{\theta}^2+\frac{d\theta}{dt}=0$
$(2\theta{t}+1)\frac{d\theta}{dt}=-{\theta}^2$
$\frac{d\theta}{dt}=\frac{-{\theta}^2}{(2\theta{t}+1)}$
at $t=0$, the value of $\theta$ and its differentiation is:
$\theta^2 (0)+\theta=1$
$\theta=1$
$d(\theta(0))=\frac{-{\theta}^2}{(2\theta{(0)}+1)}={-{\theta}^2}$
on putting the above value in $\frac{dr}{d\theta}$
$\frac{dr}{d\theta}=\frac{(2)}{3}(({1+7})^{\frac{-2}{3}}(-1)$
$\frac{dr}{d\theta}=\frac{-1}{6}$
