I have the following vector field defined over the manifold $M=\mathbb{R}^2 -(0,0)$:
$X(x,y) = \frac{x}{x^2+y^2}\partial _x + \frac{y}{x^2+y^2}\partial _y.$
I have found that $f(x,y) = \tan ^{-1} (y/x)$ is such that the vector field defined by $f$ as $[\partial_y f(x,y), -\partial_x f(x,y) ]=X_f$ is such that $X_f$ coincides with $X$.
Is this enough to say that X is a Hamiltonian vector field with respect to the standard symplectic form?
Yes, it is enough. For a vector field to be Hamiltonian all you need is to find a Hamiltonian (in your case $f$) such that your field is obtained as $$ X=Idf $$ where $I$ is the identification mapping between 1-forms and vector fields via the symplectic structure. In your case, that matrix has rows $(0,1)$ and $(-1,0)$ and the coordinates of $df$ are $(\partial_xf,\partial_yf)$. You certainly get $X$ when you perform the multiplication, therefore $X$ is a Hamiltonian field