= Solution
The <product rule> and <chain rule> first give
$$
F_x=y^2(1+2x^2y)e^{x^2y},\qquad F_y=x(2y+x^2y^2)e^{x^2y}.
$$
Differentiating the first expression with respect to $y$ yields $e^{x^2y}(2y+7x^2y^2+2x^4y^3)$; differentiating the second with respect to $x$ gives exactly the same polynomial factor. Thus
$$
\boxed{F_{xy}=F_{yx}=e^{x^2y}(2y+7x^2y^2+2x^4y^3).}
$$
This also follows from <symmetry of second derivatives>, since all derivatives of this function are continuous everywhere.
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