I would like to differentiate the function $$\ln \frac{x-y_0}{a}$$ with respect to $x$ using a substitution. Let $z = \frac{x-y_0}{a}$ and differentiate $\ln z$ with respect to $z$. Let $'$ denote differentiation with respect to $x$:
$$\frac{d}{dx}\ln z = \frac{z'}{z} = \frac{1}{z} \left(\frac{x-y_0}{a}\right)' = \frac{1}{za} = \frac{a}{a(x-y_0)} = \frac{1}{x-y_0}.$$
Is this a right solution?
On
It's the correct solution, yes, but it's not the derivative $\ln z$ by $\color{red}z$ (as you say in your answer), it's still the derivative of $\ln(z(x))$ by $\color{red}x$.
You can get the same answer in a different way by first rewriting
$$\ln\frac{x-y_0}{a} = \ln(x-y_0) - \ln a$$ and using the transformation $z=x-y_0$.
Assuming $a$ and $y_0$ are constants, yes. Your steps look good.