You are OK separating variables.
\(\displaystyle \int\frac{1}{y^{2}}dy+\int\frac{1}{2x+1}dx=0\)
\(\displaystyle \frac{-1}{y}=\frac{-ln(2x+1)}{2}+C\)
Now, we can use the IC to find C, x=4 and y=1:
\(\displaystyle \frac{-1}{1}=\frac{-ln(2(4)+1)}{2}+C\)
\(\displaystyle C=ln(3)-1\)
Using C and solving for y we get:
\(\displaystyle \boxed{y=\frac{2}{ln(2x+1)-2(ln(3)-1)}}\)