Suppose $f(x,y)\in \mathcal{C}^2$ (twice continuously differentiable, also real). When can you say $$f_x(x,y)\Biggr|_{y=0} = \partial_x\left(f(x,0)\right)$$ with the loosest possible restrictions? (I'd prefer to not have to bound the derivative, which I know works.)
Edit: (based on comment below) In particular, is it true for all $f(x,y)\in \mathcal{C}^2$ (i.e. no additional assumptions)?
Define the following
$$f(x,y) = \cases{ (x^2+y^2) \sin (\frac{1}{\sqrt{x^2+y^2}}) & (x,y) $\neq$ (0,0)\\ 0 & (x,y) = (0,0)}$$
Then $f$ does not have continuous partial derivatives at the origin; however, it is easily verified that $\partial_x f(x,0)$ is, in fact, differentiable at the origin.
I think continuity of the partial derivatives is required in order for the operations to commute.