Showing that ln$(xy)$ = ln $x$+ln $y$

We have $$f(xy) = \int_1^{xy}\dfrac{dt}t = \underbrace{\int_1^x \dfrac{dt}t+ \int_x^{xy} \dfrac{dt}t = \int_1^x \dfrac{dt}t + \int_1^y \dfrac{dz}z}_{\text{Setting $z=t/x$}} = f(x) + f(y)$$

The correct solution depends on your definition of $\ln(x)$.

One definition is: $\ln(x)$ is the unique real number satisfying $e^{\ln(x)} = x$.

Now: $e^{\ln(x) + \ln(y)} = e^{\ln(x)}e^{\ln(y)} = xy = e^{\ln(xy)}$. So we must have $\ln(x) + \ln(y) = \ln(xy)$.