Cauchy's theorem with and without Green's Theorem

Question

I know two main approaches to proof of Cauchy's theorem ($\oint_{\gamma}^{}f(z)dz=0$ for any simple connected contour $\gamma$ and any function $f$ holomorphic inside $\gamma$). First is based on Green's theorem and takes just a few lines: $$ \oint_{\gamma}^{}f(z)dz=\oint_{\gamma}(u(x,y)+iv(x,y))(dx+idy)=\oint_{\gamma}(u\enspace dx - v\enspace dy) + i\oint_{\gamma}(u\enspace dy + v\enspace dx)=\int \int_{\Gamma}(-\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y})dxdy + i \int \int_{\Gamma}(\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y})dxdy = 0$$ where the last equality follows from Cauchy-Riemann equations.
Second (Goursat theorem) proves special case when the contour is a triangle and then extrapolate result on polygons and convex contours. It's quite a classic, but for the sake of completeness I'll attach proof (from 'Complex Analysis' by John M. Howie).
Proof
Question 1: I read everywhere (including Wikipedia) that the proof based on Green's theorem is 'worse' in a sense that it assumes continuity of partial derivatives $\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}, \frac{\partial v }{\partial x}$ and $\frac{\partial v}{\partial y}$, which is unnecessary. However, to my understanding the proof of Goursat theorem uses this assumption too, because it says that function has to be holomorphic everywhere inside contour. The same author defines holomorphic function as differentiable at every point in some open set $U \subseteq \mathbb{C}$ . He also states that the sufficient condition for a function to be differentiable at point $c$ is the continuity of its partial derivatives in some vicinity of $c$, beside Cauchy-Riemann equations. (He provides an example $f(x+iy) = u(x,y) + iv(x,y)$ with $u(x,y) = \sqrt{|xy|}$ and $v(x,y) = 0$ and shows that derivative is not defined at $(0,0)$ despite C-R eq. are trivially satisfied).
So, as I see, the continuity of partials is already assumed implicitly. If the answer is that in a proof only local differentiability is used, then
Question 2: If there is a singularity inside a contour, at which point the proof of Goursat theorem breaks? If we only use the local differentiability in a point inside all the triangles, then singularities outside inner triangles should have no affect, which I know to be wrong.

Any help will be appreciated.