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Suppose that $f(t)$, $g(t)$ have a 2nd order derivative.

Let $F(x, y) := x f(\frac{y}{x}) + g(\frac{y}{x})$.

Then, $F_{xy}(x, y) = F_{yx}(x, y)$.

$F$ is not a class $C^2$ function in general.

Is there a necessary and sufficient condition to be $F_{xy}(x, y) = F_{yx}(x, y)$ when $F(x, y)$ has all the 2nd order partial derivatives?

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    Yes, its called Schwarz theorem, see https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives It states that any $C^2$ function has this property. – 2017-02-12
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    Is the following statement true? If $F$ satisfies $F_{xy} = F_{yx}$, then $F$ is a class $C^2$ function. – 2017-02-12
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    I think the theorem says that if $F$ is a class $C^2$ function, then $F_{xy} = F_{yx}$. – 2017-02-12

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