= Solution
The relevant <mixed partial derivatives> agree:
$$
\partial_y(y^2-2x)=2y=\partial_x(2xy-y).
$$
The <vector field> is <smooth> on the entire <simply connected> <plane>, so this condition establishes that it is a <conservative vector field>. We can also construct its <vector-field potential> explicitly. From $\phi_x=y^2-2x$, take an <antiderivative> in $x$:
$$
\phi=xy^2-x^2+h(y).
$$
Then $\phi_y=2xy+h'(y)=2xy-y$ requires $h'(y)=-y$. Hence
$$
\boxed{\phi(x,y)=xy^2-x^2-\frac12y^2+C,\qquad \mathbf F=\nabla\phi}.
$$
The <fundamental theorem for line integrals> checks the previous answer: $\phi(1,1)-\phi(0,0)=-1/2$.
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