Solution 1:

Hint: Your method so far is correct. What you are asking for is the integral $$\int \frac{\log\left(\frac{y}{1+y}\right)}{y+y^2}~dy.$$ The integral at first looks deceptively hard to solve but it is actually very easy if you notice it is of the form $$\int g(y)g'(y)~dy$$ for $g(y)=\log\left(\frac{y}{1+y}\right)$.

Solution 2:

Your approach is correct. As regards the integral on the left, note that $$\int \frac{1}{f(z)+f^2(z)} df = \int \left(\frac{1}{f(z)}-\frac{1}{1+f(z)}\right) df =\log\left(\frac{f(z)}{1+f(z)}\right)+c.$$ Can you take it from here?