We have a square on the plane of sides 2 from (-1,-1) to (1,1), and $P(x,y)=x^2+y^2,Q(x,y)=2x^2y$. $$ \oint_L (x^2+y^2)dx+2x^2ydy=2\int_{-1}^1(x^2+1)dx+2\int_{-1}^12ydy=\frac{16}{3}+0=\frac{16}{3} $$ as on the horizontal lines $dy=0$ and $y=\pm1$ and in the direction [1,-1] $dx<0$, and same reasoning on the verticals.
With Green's Theorem the line integral becomes $$ \int_{-1}^1\int_{-1}^1(4xy-2y)dxdy=\int_{-1}^12ydy\int_{-1}^1(2x-1)dx=0\int_{-1}^1(2x-1)dx=0. $$ I think I made a mistake in the last Green's Theorem calculation, as to another nameless theorem (can't post the proof here as it's 2 pages in my textbook): For the closed line integral inside a simply connected region $S$ $$ \oint_L P(x,y)dx+Q(x,y)dy=0\iff \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\iff 2y=4xy, $$ which is not the case. But I am not sure where my mistake is, as I verified all conditions and calculations.
Your mistake is with the $2\int_{-1}^1(x^2+1)\,dx$ term. It should be $\int_{-1}^1(x^2+1)\,dx+ \int_{1}^{-1}(x^2+1)\,dx=0$, because on the top of the square you’re going from right to left. Meaning, you didn’t take the orientation of the curve into account.
Strictly speaking, your $2\int_{-1}^12y\,dy$ term is also wrong; you didn’t take the orientation of the left side of the square into account (you’re going from top to bottom). You wouldn’t have noticed this either, because this mistake doesn’t affect the answer as the integral is zero, so it or its minus both vanish.
Finally, let me make the remark that your final 2 chain equivalence is incomplete. You’re forgetting the very important universal quantifier “for all (nice enough) closed loops $L$” in the first part.
Note, it is very much possible to have the partials be unequal, as is the case here: $2y$ is not identically equal to $4xy$, but yet there are some loops $L$ (like the boundary of this square), for which the integral vanishes. In fact, because of the equivalence, if you state, and then negate it properly, you’ll see that $2xy\neq 4xy$ identically implies there exists a loop $L$ for which the integral is non-zero (and it turns out the boundary of the square is one such loop).