To calculate the limit for $y\to\infty$ you can use L'Hospital, but not immediately as the term will be $\infty \cdot 0 - \infty$ which is not defined. You first have to convert this term to get an expression of $0/0$ so that L'Hospital can really be used.
To begin with, the functions are defined with $F,G,H$ so that the above formula is simplified to
$$
F\cdot G - H
$$
with $F\to \infty, G \to 0, H \to \infty$.
Transforming this into a formula that is allowed for L'Hospital yields
$$D = \frac{1}{\frac{1}{F\cdot H}} \cdot \left(G\cdot \frac 1H - \frac 1F\right)$$
as for $y\to\infty$ it would approach $0/0$
Let's calculate each function and its derivative for its own in terms of maximum exponents - as that is the only interesting at the end when looking at the behavior for $y\to\infty$. The behavior for large $y$ is denoted with $\sim$ which is similar to $\Theta$ in the $\mathcal{O}$ notation:
\begin{align}
F &\sim y^1 \\
G &\sim 0 \\
H &\sim y^{- \frac 12} \\
\frac{d}{dy} G &= \frac{\sqrt{1+y}-1}{\sqrt{1+y}+1}\cdot\frac{\frac{1}{2\sqrt{1+y}}(\sqrt{1+y}-1) - \frac{1}{2\sqrt{1+y}} (\sqrt{1+y}+1)}{(\sqrt{1+y}-1)^2} \\
&= \frac{-1}{(\sqrt{1+y}+1)(\sqrt{1+y}-1)(\sqrt{1+y})} \sim y^{-\frac{3}{2}}\\
\frac{d}{dy} \frac{1}{H} &=\frac{-1}{4(1+y)^{\frac{3}{2}}} \sim y^{-\frac 32} \\
\frac{d}{dy}\frac{1}{F} &= \frac{1}{-\left(y+\frac 23\right)^{2}} \sim y^{-2} \\
\frac{d}{dy}G\frac{1}{H} &\sim y^{-\frac 32} \cdot y^{-\frac 12} + 0 \cdot y^{-\frac 32} \sim y^{-2} \\
\frac{d}{dy}\frac{1}{F}\frac{1}{H} &\sim y^{-2} y^{-\frac 12} + y^1 y^{-\frac32} \sim y^{-\frac 52} + y^{-\frac 12} \sim y^{-\frac 12}
\end{align}
If we put everything together now, we conclude:
\begin{align}
D \sim \frac{1}{y^{-\frac 12}} \cdot (y^{-2} - y^{-2}) \sim y^{\frac 12} \cdot y^{-2} \sim y^{-\frac 32}\\
\end{align}
It could be possible that we get $y^{-2}-y^{-2}=0$ in the brackets, which would change the result. But if we take a look at the coefficient in front of $y^{-2}$ we get $(-1)\cdot \frac 12 = -\frac 12$ for $\frac{d}{dy}\left(G\cdot \frac 1H\right)$ (the factor of $\left(\frac{d}{dy}G\right)\frac{1}{H}$ from using the product rule as the other product has exponent $0$) and $-1$ for $\frac{d}{dy}\frac 1F$. As $-\frac 12 \neq -1$ the above difference will not be zero, therefore the exponents will remain as calculated.
So in total we have $D\sim y^{-\frac 32}$ or more mathematically:
$$
D \in \Theta\left(y^{-\frac 32}\right)
$$
Puh. That took a long time to write and think about. I hope everything is correct and clearly written. If anything is unclear or any questions arise, please ask. I would appreciate any feedback in form of comments or votes - especially as this is my first "big" answer with a more difficult topic :)
