$$G(y) = \ln \dfrac{(5y+1)^2}{\sqrt{y^2 + 1}}$$
Can we break this up like this: $$2 \ln(5y+1)\over \frac 12\ln(y^2 + 1)$$
$\ln\left(\dfrac{f(y)}{g(y)} \right) \neq \dfrac{\ln(f(y))}{\ln(g(y))}$. But make use of the fact that $$\ln\left(\dfrac{f(y)}{g(y)} \right) = \ln(f(y)) - \ln(g(y))$$ and $$\dfrac{d \ln(h(y))}{dy} = \dfrac{h'(y)}{h(y)}$$
Move your mouse over the gray area for complete solution.
$$G(y) = \ln\left(\dfrac{(5y+1)^2}{\sqrt{y^2+1}} \right) = \ln((5y+1)^2) - \ln(\sqrt{y^2+1}) = 2 \ln(5y+1) - \dfrac12 \ln(y^2+1)$$Hence, \begin{align}\dfrac{dG(y)}{dy} & = 2 \times \dfrac1{5y+1} \times 5 - \dfrac12 \times \dfrac1{y^2+1} \times (2y) \\& = \dfrac{10}{5y+1} - \dfrac{y}{y^2+1}\\ & = \dfrac{10y^2 + 10 - 5y^2 - y}{(5y+1)(y^2+1)}\\ & = \dfrac{5y^2 - y +10}{(5y+1)(y^2+1)} \end{align}
No: is $\ln\frac{a}b$ equal to $\frac{\ln a}{\ln b}$, or is it equal to some other combination of $\ln a$ and $\ln b$? (You did handle the exponents correctly, though.)