How to prove relationship between mutual information and differential entropy?
Your derivation is correct. Your second integral can also be written as

$$
\int_{x,y}f(x,y)\log_2(f(y))dxdy=\int_y\log_2(f(y))\left(\int_xf(x,y)dx\right) dy
$$

where the integral in parenthesis is equal to 

$$
\int_xf(x,y)dx=\int_xf(x|y)f(y)dx=f(y)\int_xf(x|y)dx
$$

You can understand this integral in two ways:


Like your idea: $\int_xf(x|y)dx=1$, which means that regardless of what value $y$ has, $x$ must have some value, so the integral over all possibilities is 1. You can understand that also by seeing $f(x|y)$ as some distribution for $x$ (which is parametrized by $y$). And, the integral over the support of a distribution needs to be $1$. 
$\int_xf(x,y)dx=f(y)$ is the marginalization of $f(x,y)$ regarding the variable $x$. With the integration, you remove the influence (i.e. marginalize out) of the variable $x$ from the probability density.