Step 1: Take logarithm of both sides.
Given $y=\left(\dfrac{a+x}{b+x}\right)^{a+b+2x}$, take natural log: $\ln y=(a+b+2x)\,\big[\ln(a+x)-\ln(b+x)\big]$.
Step 2: Differentiate using the product rule.
$\dfrac{1}{y}\dfrac{dy}{dx}=2\big[\ln(a+x)-\ln(b+x)\big]+(a+b+2x)\left[\dfrac{1}{a+x}-\dfrac{1}{b+x}\right]$.
Step 3: Put $x=0$ in the bracket terms.
The first bracket becomes $\ln a-\ln b=\ln\dfrac{a}{b}$, so $2\ln\dfrac{a}{b}$. For the second, $a+b+2x$ becomes $a+b$, and $\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{b-a}{ab}$.
Step 4: Combine the right side at $x=0$.
$\left.\dfrac{1}{y}\dfrac{dy}{dx}\right|_{0}=2\ln\dfrac{a}{b}+(a+b)\cdot\dfrac{b-a}{ab}=2\ln\dfrac{a}{b}+\dfrac{b^2-a^2}{ab}$, since $(a+b)(b-a)=b^2-a^2$.
Step 5: Find $y$ at $x=0$.
At $x=0$, $y=\left(\dfrac{a}{b}\right)^{a+b}$.
Step 6: Multiply to get the derivative.
$\left.\dfrac{dy}{dx}\right|_{0}=y\cdot\left(2\ln\dfrac{a}{b}+\dfrac{b^2-a^2}{ab}\right)=\left(2\ln\dfrac{a}{b}+\dfrac{b^2-a^2}{ab}\right)\left(\dfrac{a}{b}\right)^{a+b}$. \[ \boxed{\left(2\log\dfrac{a}{b}+\dfrac{b^2-a^2}{ab}\right)\left(\dfrac{a}{b}\right)^{a+b}} \]