I don't see a square, only a cube. Does this have something to do with the absolute value?

Yes, squaring both sides, we can write:

\(\displaystyle y^2=(2x-\tan(x))^6\)

Now we can implicitly differentiate:

\(\displaystyle 2yy'=6(2x-\tan(x))^5(2-\sec^2(x))\)

\(\displaystyle y'=\frac{3(2x-\tan(x))^5(2-\sec^2(x))}{|2x-\tan(x)|^3}=3(2x-\tan(x))|2x-\tan(x)|(2-\sec^2(x))\)