= Solution
The field is smooth on the <simply connected> domain $\mathbb R^3$. Its <curl> is $\nabla\times\mathbf F=(2\lambda-2,2\mu,0)$, so a necessary and sufficient condition for a <conservative vector field> is
$$
\boxed{\lambda=1,\qquad\mu=0.}
$$
At those values it is explicitly the <gradient> of $\Phi=xy-x^2/2-yz+z^2/2$. Equality of the two previous path <integrals> would impose only $\lambda-\mu=1$, which is weaker: <coinciding path integrals need not imply a conservative field>. For example $\lambda=2,\mu=1$ makes those two <integrals> equal but leaves nonzero curl.
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