Let $u(x,y)$ be some harmonic function with continuous partials. Let $v(x,y)= \int_{(a,b)}^{(x,y)}[-u_s(s,t) ds + u_s(s,t) dt]$. Find $v_y$ at some point $(x_0,y_0)$.
I don't think anything like Leibniz' rule works here, so I think all we are left with is using the definition of the partial derivative. $v_y(x_0,y_0)= \lim_{h \to 0} \frac{1}{h}[v(x_0,y_0+h)-v(x_0,y_0)] = \lim_{h \to 0} \frac{1}{h}[\int_{(x_0,y_0)}^{(x_0,y_0+h)} -u_ydx+u_x dy]= \lim_{h \to 0} \frac{1}{h} [\int_{(x_0,y_0)}^{(x_0,y_0+h)} u_x(x_0,y_0+th) h \ dt] = \lim_{h \to 0} [\int_{(x_0,y_0)}^{(x_0,y_0+h)} u_x(x_0,y_0+th) h \ dt] = u_x(x_0,y_0). $
I used path-independence of $v$ ( by Greene's theorem, noting partials are $C^1$) in the 2nd line, choosing the straight line path from $y_0$ to $y_0 +h$.
I am not at all sure about this argument, so if anyone could tell me where I've gone wrong that would be great!