I have a function
$$F(x)=\left[ a\ln\frac{y(x)}{b}-c\ln\frac{y(x)}{d} \right]^{-2},$$
where $a$, $b$, $c$, $d$ are constant.
If the function $y(x)$ can be written as it mean value plus a disturbance, $y(x)=y_0+\epsilon * dy(x)$, where $y_0$ is a constant and $\epsilon\ll1$. How to linearize the function $F(x)$ with respect to $y_0$? In the linearization, I need to eliminate any variable preceded by $\epsilon$ with power higher than 1 and any product of two variables preceded by $\epsilon$. For example, it is easy to linearize the function $\ln\frac{y(x)}{b}$:
$$\ln\frac{y(x)}{b}=\ln(y_0+\epsilon dy)-\ln b=\ln\left[y_0(1+\epsilon \frac{dy}{y_0}) \right]-\ln b=\ln\frac{y_0}{b}+\ln\left(1+\epsilon\frac{dy}{y_0} \right)=\ln\frac{y_0}{b}+\epsilon \frac{dy}{y_0}+O(\epsilon^2).$$
Thank you very much!
First of all, Welcome to the site !
As uou take it, the problem you are facing is the composition of Taylor series and I suggest you to do it from inside to outside.
First, assuming $b>0$, $d>0$, $y(x)>0$, rewrite $$a \log \left(\frac{y}{b}\right)-c \log \left(\frac{y}{d}\right)=(a-c)\log (y)+ [c \log(d)-a \log(b)]$$ that is to say $$\alpha \log (y)+\beta \quad \text{with}\quad\alpha=(a-c)\quad \text{and}\quad \beta=c \log(d)-a \log(b)$$
Now, using Taylor, just as you wrote, $$y=y_0+y'_0 \,\epsilon\implies \log(y)=\log (y_0)+\frac{y'_0 }{y_0}\epsilon+O\left(\epsilon ^2\right)$$ So, the quantity in brackets is
$$(\beta +\alpha \log (y_0))+\frac{\alpha y'_0 }{y_0}\epsilon +O\left(\epsilon ^2\right)$$ Square it $$(\beta +\alpha \log (y_0))^2+\frac{2 \alpha y'_0 (\beta +\alpha \log (y_0))}{y_0}\epsilon+O\left(\epsilon ^2\right)$$ and now, use the long division
$$F(x)=\frac{1}{(\beta +\alpha \log (y_0))^2}-\frac{2 \alpha y'_0}{y_0 (\beta +\alpha \log (y_0))^3}\epsilon+O\left(\epsilon ^2\right)$$
For sure, you need to replace $y_0$ by $y(x_0)$, $y'_0$ by $y'(x_0)$ and $\epsilon$ by $(x-x_0)$.
Edit
But, using the first principles, linearizing a function at $x_0$ is just the equation of the tangent line at this point, that is to say
$$F(x)=F(x_0)+F'(x_0)\, (x-x_0)$$ $$F(x)=\Bigg[a \log \left(\frac{y(x_0)}{b}\right)-c \log \left(\frac{y(x_0)}{d}\right)\Bigg]^{-2}-$$ $$2(a-c) \Bigg[a \log \left(\frac{y(x_0)}{b}\right)-c \log \left(\frac{y(x_0)}{d}\right)\Bigg]^{-3}\,\,\frac{y'(x_0)}{y(x_0)}\,\, (x-x_0)$$ which is much faster and does not require any $O((x-x_0)^2)$ consideration.