Unsure how to treat y in this derivative/log problem Need to find the derivative of $h(y)= \ln(y^2 \cos y)$
Treating it like a normal variable like an x isn't working for me, the way we used y's in earlier problems where you get a y' in there doesn't seem right, so I'm not quite sure here. 

If $h(y) = \ln(y^2 \cos y)=2 \ln y + \ln(\cos y)$, then
$$
h'(y) = \frac{2}{y} + \frac{-\sin y}{\cos y} = \frac{2}{y} - \tan y
$$
Stella is right.  In this problem, $y$ is an independent variable.  You may have worked other problems where $y$ was dependent on $x$, but this is not the case here.
