Let $$f(x,y) := xy\exp\left(-\frac{1}{x^2+y^2}\right),$$ if $(x,y)\neq (0,0)$ and $f(0,0):=0$.
I read the claim that $f$ is
(a) separately real analytic on $\mathbb{R}\times\mathbb{R}$ (i.e. for each fixed $y$ the map $x\mapsto f(x,y)$ is real analytic and for each fixed $x$ the map $y\mapsto f(x,y)$ is real analytic),
(b) $C^\infty$ on $\mathbb{R}\times\mathbb{R}$,
(c) not jointly real analytic near $(0,0)$.
I see how to prove (b), but am not sure how to go about (a) and (c). Any hints would be much appreciated.
I gave a brief explanation of (c) elsewhere, but here is a different argument. Suppose $f(x,y) =\sum_{m,n=0}^\infty a_{m,n}x^m y^n$ in a neighborhood of $(0,0)$. Then $$f(x,x) = \sum_{m,n=0}^\infty a_{m,n}x^{m+n} = \sum_{k=0}^\infty c_k x^k$$ where $c_k=\sum_{m+n=k}a_{m,n}$. (Simply put, the restriction of a real-analytic function to a line is real-analytic.) Take the smallest $k$ such that $c_k\ne 0$; it must exist since $f$ is not identically zero in any neighborhood of $0$. Then $$ c_k = \lim_{x\to 0}x^{-k} f(x,x) =\lim_{x\to 0}x^{2-k}\exp(-1/x^2) = 0, $$ a contradiction. (The last limit can be evaluated, for example, by substituting $u=\exp(1/x^2)$ and using L'Hôpital's rule.)