 Solution

Take the projective line $X={\mathbb{P}}_k^1$ and the coherent sheaves $\mathcal{F}={\mathcal{O}}_X$ and $\mathcal{G}={\mathcal{O}}_X(-1)$, where the latter is a twisting sheaf on projective space. Both are line bundles, so each stalk is a free rank-one ${\mathcal{O}}_{X,x}$-module and therefore ${\mathcal{F}}_x\simeq {\mathcal{G}}_x$ at every point.

They are nevertheless not isomorphic. On the standard charts with coordinates $t$ and $t^{-1}$, choose frames $e_0,e_1$ of $\mathcal{O}(-1)$ with $e_1=t^{-1}{e}_0$. A global section would give polynomials $a(t)$ and $b(t^{-1})$ satisfyingThe left side has only nonnegative powers of $t$, and the right side only strictly negative powers. Thus both vanish, giving $\Gamma (X,\mathcal{O}(-1))=0$. For the structure sheaf, the corresponding equality is $a(t)=b(t^{-1})$, whose solutions are precisely the constants. HenceA morphism of sheaves which is an isomorphism on all stalks would be an isomorphism; this example only asserts the existence of separate stalk isomorphisms, not that they come from one morphism of sheaves.