Given that $\rho=\tan(\rho+\phi)$, how to find $\frac{d^3 \rho}{d\phi^3}$?
Answer is: $$-\frac{2(5+8\rho^2+3\rho^4)}{\rho^8}$$
I start from $$\rho=\tan(\rho+\phi) \tag{1} $$ and use $$\tan(a+b)=\frac{\tan a +\tan b}{1-\tan a \tan b}$$ On applying this to eq (1) :
$$\tan \phi +\tan \rho=\rho(1-\tan\phi \tan\rho)$$ On further substitution i got:
$$\tan\phi+\tan \rho=\rho-\rho \tan\phi \tan\rho$$please can someone help me out here.
Hint
I think that the easier way is implicit differentiation. Start with $$F=\rho-\tan(\rho+\phi)=0$$ from which $$F'_{\rho}=1-\sec ^2(\rho +\phi )$$ $$F'_{\phi}=-\sec ^2(\rho +\phi )$$ So, now $$\frac{d\rho}{d\phi}=\frac{\sec ^2(\rho +\phi )}{1-\sec ^2(\rho +\phi )}=-1-\cot^2(\rho+\phi)=-1-\frac{1}{\rho^2}$$
I am sure that you can take from here.
Added later
In the first step, what was found is that $$\rho '(\phi )=-1-\frac{1}{\rho(\phi)^2}$$ Continuing the differentiation with respect to $\phi$ $$\rho ''(\phi )=\frac{2 \rho '(\phi )}{\rho (\phi )^3}$$ $$\rho ^{(3)}(\phi )=\frac{2 \rho ''(\phi )}{\rho (\phi )^3}-\frac{6 \rho '(\phi )^2}{\rho (\phi )^4}$$ $$\rho^{(4)}(\phi )=\frac{2 \rho ^{(3)}(\phi )}{\rho (\phi )^3}+\frac{24 \rho '(\phi )^3}{\rho (\phi )^5}-\frac{18 \rho '(\phi ) \rho ''(\phi )}{\rho (\phi )^4}$$ and so on