# Given first order partial derivatives exist, is it always possible that mixed partial exist?

If two-variable function $$\ f(x,y)$$ is partialy differentiable respect to both variables ; if both $$\frac{\partial f}{\partial x}$$ and $$\frac{\partial f}{\partial y}$$ exist, is it always true the second mixed partial $$\frac{\partial^2 f}{\partial x \partial y}$$ exist?

If it's not the general case, what could be the weakest condition that fit?

Well using my intuition $$\ f_x$$ and $$\ f_y$$ being continuous is the best I could imagine, but I guess that's not all? Please enlighten me.

I'm currently learning calculus by Stewart calculus ed.8

If exact answer requires rigorous understanding of math concepts Which books/texts do I need to look for?

Thanks

Try $$f(x,y)=e^{1/(x^2+y^2)} x y, \qquad f(0,0)=0$$ Away from $$(0,0)$$ it is smooth, but at $$(0,0)$$ only the partial derivatives in directions $$(1,0)$$ and $$(0,1)$$ exist, not in the direction $$(1,1)$$. The $$\frac{\partial^2 f}{\partial x \partial y}$$ second derivative doesn't exist neither.
$$g(x,y)=xy\sin(\frac1{x^2+y^2})$$ It is smooth away from $$(0,0)$$.
$$g(x,y)=O(x^2+y^2)$$ so the derivatives in every directions exist and vanish at $$(0,0)$$, but $$(\partial_x g)(0,y)=y\sin(1/y^2)$$ which is not differentiable at $$y=0$$ so that $$(\partial_y (\partial_x g))(0,0)$$ doesn't exist.