Well, it follows directly from the definition that $g$ and all its partial derivatives are uniformly continuous on the closed ball $\{(x,y) \in \mathbb{R}^2 | x^2+y^2 <= 1, y <= 0\}$. It also follows immediatly that $g \in C^2(\{(x,y) \in \mathbb{R}^2 | x^2+y^2 < 1, y \neq 0\})$. So what remains to shows is that the partial derivatives up to the second order exists at $y=0$.
For the first-order partial derivative $\partial_y g$ you have $g_y(x,0) = \lim_{\epsilon \rightarrow 0^+} \frac{g(x,\epsilon)-g(x,-\epsilon)}{2\epsilon}$. Since $-g(x,-\epsilon)=g(x,\epsilon)=f(x,\epsilon)$ this is the same as $\lim_{\epsilon \rightarrow 0^+} \frac{f(x,\epsilon)}{\epsilon} = f_y(x,0)$. A similar approach should work for the other partial derivatives.
To convince yourself that $f_y(x,0)$ (i.e. the continuous extension of $f_y$) coincides with the one-sided derivative of $f$ at $(x,0)$, write the former out as a double limit $ f_y(x,0) = \lim_{y \rightarrow 0^+} \lim_{\epsilon \rightarrow 0^+} \frac{f(x,y+\epsilon)-f(x,y)}{\epsilon}. $ Note that since $f_y$ is bounded, the inner limit converges uniformly with respect to $y$. You may thus swap the limits, whith shows that $ f_y(x,0) = \lim_{\epsilon \rightarrow 0^+} \frac{f(x,\epsilon)}{\epsilon}. $